The new volume of the balloon at a temperature of 325 K is approximately 0.59 L.
We use the combined gas law to solve this problem, which relates the pressure, volume, and temperature of the gas;
P₁V₁/T₁ = P₂V₂/T₂
where P is pressure, V is volume, and T temperature.
We know the initial volume (V₁) is 0.50 L and the initial temperature (T₁) is 273 K. We also know that the pressure remains constant, so we can set P₁ = P₂. Finally, we need to find V₂, the new volume at a temperature of T₂ = 325 K.
Substituting these values into the equation, we get;
P₁V₁/T₁ = P₂V₂/T₂
P₁ (0.50 L)/(273 K) = P₂ V₂/(325 K)
Simplifying, we get;
V₂ = (P₁/P₂) × (T₂/T₁) × V₁
We don't know the pressure of the gas, but we know it remains constant, so we can cancel it out;
V₂ = (T₂/T₁) × V₁
Plugging in the numbers, we get:
V₂ = (325 K/273 K) × 0.50 L
V₂ = 0.59 L
Therefore, the new volume of the balloon is 0.59 L.
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A solution contains 37.5 grams of calcium carbonate (caco3) in 500 ml of
water. what is the concentration of this solution?
The concentration of the solution is 0.748 M.
To find the concentration of the solution, we need to calculate the number of moles of calcium carbonate present in the solution.
First, we need to determine the molecular weight of calcium carbonate ([tex]CaCO3[/tex]).
[tex]CaCO3[/tex] = 1 x Ca + 1 x C + 3 x O
= 40.08 g/mol + 12.01 g/mol + (3 x 16.00 g/mol)
= 100.09 g/mol
Next, we can use the formula:
concentration (in mol/L) = moles of solute / volume of solution (in L)
We have 37.5 grams of calcium carbonate in 500 ml of water. To convert ml to L, we divide by 1000:
volume of solution = 500 ml / 1000 ml/L = 0.5 L
moles of calcium carbonate = mass / molecular weight
= 37.5 g / 100.09 g/mol
= 0.374 moles
Therefore, the concentration of the solution is:
concentration = 0.374 moles / 0.5 L
= 0.748 M
The concentration of the solution is 0.748 M.
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HEAT
INTRODUCTION
Heat is a measure of the energy in a system. The transfer of energy is always from the system with more energy to the system with less energy. This lab has two distinct parts. In the first part, you will examine what happens to a gas when the temperature is changed. In the second part, you will use the idea of energy transfer to move water. You will need to be familiar with the ideas of phases (solid, liquid and gas), what specific heat is, and how to calculate joules. Please see pages 93-94, 99-101, and 106-110 in your textbook.
MATERIALS
1 small mouth (or small neck) bottle… a soda bottle should work
1 coin (dime or penny – must cover completely mouth of bottle)
1 large container to submerge at least ½ the bottle (sink, tub, bowl, etc.)
Enough cold water to submerge ½ the bottle
Measuring cups
Food coloring – in kit
4 cups water
1 large bowl to hold water – a clear glass one works best
1 small glass that will extend above water level when in bowl
Saran wrap/cling film – enough to cover bowl
1 small object (example: pebble, coin, marble)
Sunny days (3-4)
Lab 11 - Heat
Page 1 | 4
PART#1: Magic Coin?
Procedure:
Fill selected container with some cold water.
Place the bottle and coin in the bowl of water to chill them. The bottle must be submerged upside down. Submerge at least the neck of the bottle but if you have no “coin activity” on step four, repeat this step with either a greater amount of submersion or submerge the bottle for a greater amount of time.
Place the coin on the top of the bottle. There should be an airtight seal when you place the coin on the top of the bottle.
Wrap your hands around the bottle and wait for several seconds to a minute.
When you believe that the bottle is warmer than room temperature, allow the bottle to cool with the coin in place. Answer the following questions based on your observations.
Questions:
Approximately how long did you submerge the bottle in step #2?
What happened during step #4?
What happened during step #5?
Explain what is happening to the molecules to create the “coin activity”.
PART#2: Distillation
Procedure:
Add the water to the bowl.
Stir in the food coloring until it is distributed equally.
Place the empty glass (small) in the middle of the large bowl so that none of the
Lab 11 - Heat
Page 2 | 4
colored water can get into the glass. The glass must be short enough that it does not extend beyond the rim of the glass bowl.
Note: If the glass bowl is not working because the small empty glass is not stable, a stock pot/dutch oven (with a flat bottom) will work but it will need to be left alone for a little more time.
Cover the large bowl completely with the saran wrap so that no air can pass through.
Add the small object on the saran wrap so that the saran wrap dips in over the small empty glass but does not cause the saran wrap to slip off the lip of the bowl. Use a smaller pebble or coin if the first one is too heavy.
Leave the bowl in the sunlight for a few days and watch to see what happens.
Remove the small glass and measure the amount of water in it with the measuring cups (estimating to the nearest 1/8 cup). Contact me immediately if the amount of water in the small glass is less than 1/8 cup.
Questions:
How is the water in the large bowl different from the water in the small glass?
Describe step by step what happened to the water that is now in the small glass in terms of heating/cooling, phase changes, etc. (Hint: there is more than one step required)
How many cups of water (to the nearest 1/8 cup) are in the small glass?
How many grams of water did you collect?
The relationship between cups and grams is: 1 cup = 236 grams
How many calories are needed to heat the water?
Assume the following information:
The original temperature of the water in the large bowl was 25 °C.
The temperature of a molecule that changes from liquid to gas is 100 °C.
The specific heat of water is 1.00 cal/g·°C
Lab 11 - Heat
Page 3 | 4
You will need the equation for specific heat (equation 4.4)
How many calories are needed to evaporate the water?
The latent heat of vaporization of water is 540.0 cal/g
You will need equation 4.6 in the textbook.
How many calories (total) are needed to “move” the water from the large bowl to the small glass?
Notes: Ignore the amount of water that was not “moved” The water molecules must warm AND change state
Answer:
Hello! This lab is all about heat, which is a measure of energy in a system. In the first part, we'll be examining what happens to a gas when the temperature changes. For this part, you will need a small mouth bottle, a coin, a large container, cold water, and measuring cups. In the second part, we'll be using the idea of energy transfer to move water. For this part, you will need food coloring, water, a large bowl, a small glass, cling film, a small object, and sunny days. Follow the procedures carefully and answer the questions provided to understand the concepts of heat and energy transfer. Don't hesitate to reach out if you have any questions!
C water = 1 cal/g ℃
How is the (Delta)Hfusion used to calculate volume of liquid frozen that produces 1 kJ of energy?
Delta Hfusion is a term used in thermodynamics to refer to the amount of energy that is required to convert a substance from its solid state to its liquid state, or vice versa, at a constant pressure. This energy is typically expressed in terms of Joules per unit mass, such as J/g or kJ/kg.
To calculate the volume of liquid that is frozen, we first need to determine the amount of mass that is required to produce 1 kJ of energy. This can be calculated using the equation:
q = m * Delta Hfusion
where q is the amount of energy produced (in J), m is the mass of the substance being frozen (in kg), and Delta Hfusion is the amount of energy required to freeze the substance (in J/kg). Rearranging this equation to solve for m, we get:
m = q / Delta Hfusion
Substituting the values of q = 1 kJ and Delta Hfusion (which is a known value for the substance being frozen), we can calculate the mass of the substance required to produce 1 kJ of energy. Once we know the mass, we can use the density of the substance to calculate the volume of liquid that is frozen.
For example, let's say we are trying to freeze water to produce 1 kJ of energy. The Delta Hfusion of water is 333.6 kJ/kg. Using the equation above, we can calculate the mass of water required to produce 1 kJ of energy:
m = (1 kJ) / (333.6 kJ/kg) = 0.003 kg
Next, we can use the density of water (which is approximately 1000 kg/m^3) to calculate the volume of water that is frozen:
Volume = mass / density = 0.003 kg / 1000 kg/m^3 = 0.000003 m^3
So, the volume of water that is frozen to produce 1 kJ of energy is approximately 0.000003 cubic meters, or 3 milliliters.In summary, we can use the Delta Hfusion of a substance, along with its density, to calculate the volume of liquid that is frozen to produce a certain amount of energy.
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Use the Beer’s law plot and best fit line to determine the concentrations for samples: M21050-1 0. 359
M21050-2 0. 356
M21050-3 0. 339
M21050-4 0. 376
M21050-5 0. 522
Beer's law establishes a connection between a substance's concentration in a solution and its absorbance at a certain wavelength.
By charting the absorbance vs concentration of each solution, a Beer's law plot is created in order to calculate concentrations of a series of copper(II) sulfate solutions with known absorbances at a set wavelength. The resulting graph should be a straight line that the least-squares approach can fit with a linear equation. The molar absorptivity is represented by slope of the best-fit line, and the absorbance at zero concentration is represented by the y-intercept. By measuring the absorbance of the unknown copper(II) sulfate solutions and solving for concentration using the equation, the concentrations of unknown solutions can be found.
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--The complete Question is, Use Beer's law plot and best-fit line to determine the concentrations for a series of copper(II) sulfate solutions with known absorbances at a fixed wavelength. --
11. 2H202 (1) - 2H20 (1) + 02(g)
Drake asked Theo why the decomposition of hydrogen peroxide, H202, loses mass, especially when there are more molecules on the product side. Theo explains that it is because they decomposed the product. He says that decomposing the product destroys the original substance. To further prove his point, he explains that in nature, decomposition occurs when dead organic matter is destroyed by fungi: without this, the world would be littered with dead things. What, if anything, is wrong with this conversation of
what happened in the reaction? Justify your answer.
A few errors about hydrogen peroxide's breakdown can be found throughout the discourse. Instead of being destroyed, the product is transformed into water and oxygen.
What happens when water and oxygen are formed from hydrogen peroxide?Catalase enzymes are found in both plants and animals, and they catalyse the conversion of hydrogen peroxide into water and oxygen. Water and oxygen are naturally formed from hydrogen peroxide, although the process is extremely slow.
How can you gauge how quickly hydrogen peroxide breaks down?Time how long it takes a disc of filter paper to rise a specified distance in a test tube containing hydrogen peroxide solution as one method of determining the rate.
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How many joules of energy do you release or lose to turn 460. g of nh3 from a liquid back to a solid?
The energy required to change 460 g of NH₃ from a liquid to a solid is roughly 152.86 kJ.
To calculate the energy released or lost when turning 460 g of NH₃ (ammonia) from a liquid to a solid, we need to determine the amount of heat energy involved in the phase transition. This can be done using the heat of fusion, which is the amount of heat energy required to convert a substance from a solid to a liquid or vice versa.
The heat of fusion of NH₃ is approximately 5.65 kJ/mol. We need to convert the mass of NH₃ to moles to use this value. The molar mass of NH₃ is 17.03 g/mol.
First, we calculate the number of moles of NH₃:
moles = mass / molar mass
moles = 460 g / 17.03 g/mol
moles ≈ 27.01 mol
Next, we calculate the energy released or lost:
energy = moles × heat of fusion
energy = 27.01 mol × 5.65 kJ/mol
energy ≈ 152.86 kJ
Therefore, approximately 152.86 kJ of energy would be released or lost when converting 460 g of NH₃ from a liquid to a solid.
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How many moles of N2 are in a flask with a volume of 250mL at a pressure of 300. 0kPa and a temperature of 300. 0K?
There are approximately 0.0364 moles of N2 in the flask.
To calculate the number of moles of N2 in a flask with a volume of 250mL at a pressure of 300.0kPa and a temperature of 300.0K, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, we need to convert the volume from mL to L by dividing it by 1000: 250mL ÷ 1000 = 0.25L.
Next, we need to convert the pressure from kPa to atm by dividing it by 101.3 (which is the conversion factor between kPa and atm): 300.0kPa ÷ 101.3 = 2.96atm.
Now we can plug in the values and solve for n: n = (PV) / (RT) = (2.96atm x 0.25L) / (0.08206 L·atm/mol·K x 300.0K) = 0.0364 moles of N2.
Therefore, there are approximately 0.0364 moles of N2 in the flask.
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Which molecule has the shortest carbon-oxygen bond length?
A. CH3COOH
B. CH3CH2OH
C. CO₂
D. CO
What set of coefficients will balance the chemical equation below:
___H2SO4 (aq) + ___NH4OH (aq) ---> ___H2O (l) + ___(NH4)2SO4 (aq)
A. 2,2,1,2
B. 1,2,2,1
C. 1,1,2,2
D. 1,3,2,1
The set of coefficients that will balance the chemical equation are: B.) 1, 2, 2, 1
What is meant by chemical reaction?Chemical reactions are processes that cause one set of chemical elements to change into another set of chemical elements. During chemical reaction, atoms are rearranged, bonds between atoms are broken and formed and then new compounds or molecules are produced.
Chemical reactions can be represented using the chemical equations, that show reactants and products.
The balanced chemical equation for the given reaction is: H₂SO₄ (aq) + 2NH₄OH (aq) ---> 2H₂O (l) + (NH₄)2SO₄ (aq)
Therefore, the set of coefficients that will balance the chemical equation are: 1, 2, 2, 1.
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2n2o5 (g) = 4no2 (g) + o2(g)
if the rate of decomposition of n2o5 at a particular instant in a reaction vessel is 4.2 x 10-7 m/s,
what is the rate of appearance of a) no2 b) o2
The rate of appearance of [tex]NO_{2}[/tex] is 8.4 x [tex]10^{-7}[/tex] m/s and the rate of appearance of [tex]O_{2}[/tex] is 2.1 x [tex]10^{-7}[/tex] m/s.
Given the reaction: 2[tex]N_{2}O_{5}[/tex](g) → 4[tex]NO_{2}[/tex](g) + [tex]O_{2}[/tex](g)
The rate of decomposition of [tex]N_{2}O_{5}[/tex] is 4.2 x [tex]10^{-7}[/tex] m/s.
a) To find the rate of appearance of [tex]NO_{2}[/tex], we will look at the stoichiometric coefficients in the balanced reaction. For every 2 moles of [tex]N_{2}O_{5}[/tex] decomposed, 4 moles of [tex]NO_{2}[/tex] are produced. So, the ratio is 4:2, which simplifies to 2:1.
Rate of appearance of [tex]NO_{2}[/tex] = (Rate of decomposition of [tex]N_{2}O_{5}[/tex]) x (2/1)
Rate of appearance of [tex]NO_{2}[/tex] = (4.2 x [tex]10^{-7}[/tex] m/s) x 2
Rate of appearance of [tex]NO_{2}[/tex] = 8.4 x [tex]10^{-7}[/tex] m/s
b) For the rate of appearance of [tex]O_{2}[/tex], we will again look at the stoichiometric coefficients. For every 2 moles of [tex]N_{2}O_{5}[/tex] decomposed, 1 mole of [tex]O_{2}[/tex] is produced. The ratio is 1:2.
Rate of appearance of [tex]O_{2}[/tex] = (Rate of decomposition of [tex]N_{2}O_{5}[/tex] ) x (1/2)
Rate of appearance of [tex]O_{2}[/tex] = (4.2 x [tex]10^{-7}[/tex] m/s) x 1/2
Rate of appearance of [tex]O_{2}[/tex] = 2.1 x [tex]10^{-7}[/tex] m/s
Thus, the rate of appearance of [tex]NO_{2}[/tex] is 8.4 x [tex]10^{-7}[/tex] m/s and the rate of appearance of [tex]O_{2}[/tex] is 2.1 x [tex]10^{-7}[/tex] m/s.
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How many grams of CaCO3 are produced when 98. 2 grams of CaO are reacted with an excess of Co2 according to the equation provided? CaO+CO2-->CaCO3
175.17 grams of CaCO₃ are produced when 98.2 grams of CaO are reacted with an excess of CO₂ according to the given equation.
To solve this problem, we need to use stoichiometry which deals with the quantitative relationships between reactants and products in chemical reactions.
The balanced chemical equation for the reaction is:
CaO + CO₂ → CaCO₃
This equation tells us that for every 1 mole of CaO and 1 mole of CO₂ that react, we get 1 mole of CaCO₃.
We are given the mass of CaO that is used in the reaction. To calculate the mass of CaCO₃ that is produced, we need to use stoichiometry and the molar mass of CaCO₃.
The molar mass of CaCO₃ is the sum of the atomic masses of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). Using the values from the periodic table, we can calculate the molar mass of CaCO₃ as:
molar mass of CaCO₃ = 1 × atomic mass of Ca + 1 × atomic mass of C + 3 × atomic mass of O
= 1 × 40.08 g/mol + 1 × 12.01 g/mol + 3 × 16.00 g/mol
= 100.09 g/mol
To calculate the number of moles of CaO that reacted, we can use the following equation:
n = m/M
where n is the number of moles of CaO, m is the mass of CaO, and M is the molar mass of CaO.
Using the given values, we get:
n = 98.2 g / 56.08 g/mol = 1.749 mol
This is the number of moles of CaO that reacted in the reaction.
Since the reaction is 1:1, meaning that one mole of CaO reacts with one mole of CO₂ to produce one mole of CaCO₃, we know that the number of moles of CaCO₃ produced is also 1.749 mol.
Finally, to calculate the mass of CaCO₃ produced, we can use the following equation:
m = n × M
where m is the mass of CaCO₃ produced, n is the number of moles of CaCO₃ produced, and M is the molar mass of CaCO₃.
Using the given values, we get:
m = 1.749 mol × 100.09 g/mol = 175.17 g
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What is the descrepancy gap between customers& expectation and perception towards service quality of front office staff/ night auditor
The discrepancy gap between customer expectations and perceptions towards service quality of front office staff/night auditor is commonly referred to as the "service gap."
This gap arises when customers have certain expectations regarding the level of service they will receive, but their actual perceptions of the service fall short of those expectations.
The service gap can be caused by a variety of factors, including inadequate training of front office staff, poor communication between staff and customers, inconsistencies in service delivery, and failure to meet customer needs and preferences.
To reduce the service gap, it is important for organizations to have a clear understanding of customer expectations and to ensure that their service delivery meets or exceeds those expectations.
This may involve implementing better training programs for front office staff, improving communication with customers, and implementing systems for monitoring and measuring customer satisfaction.
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3. Amari wants to set up a tent. He needs four 8 ft ropes. The package of ropes he bought from the store is 28 yards long. After setting up the tent, does Amari have any rope left over? If so, how much?
Answer: yes there is rope left over. 52ft of rope.
Explanation:
there are 3 ft in a yard so Amari has
3ft x 28yards = 84ft of rope
he needs 4 x 8ft = 32ft of rope
subtract what he needs from what he has to find out if he has enough and how much extra.
84ft - 32ft = 52ft of extra rope
Give the correct IUPAC name for the compound. Based on the given name, write the correct IUPAC structure for the compound
According to the question Given Name: 2-methylbutane IUPAC Name: 2-methylbutane Structure: [tex]CH_3CH_2CH(CH_3)CH_3.[/tex]
What is structure?Structure is the arrangement or organization of parts or elements in a material, system, or entity. It is essential in understanding how something is composed and how it functions. Examples of structures include the skeletal system of the human body, the structure of a computer program, the structure of a book, or the structure of a business. Structures can be physical or abstract, and are usually determined by the purpose of the material, system, or entity. For example, a bridge is a physical structure designed to support the movement of people, goods, and vehicles across a body of water. A book is an example of an abstract structure, with a specific beginning, middle, and end. The structure of a business might include the organizational hierarchy, the roles and responsibilities of each employee, and the different departments. Knowing the structure of something can help people understand how it works and how to interact with it.
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HALIDES 1. Give the definition for oxidation and reduction. (0. 4 pts) 2. If we were to mix a silver nitrate solution with the following halide containing salts, which one would produce a precipitate. CaF2, MgCl2, LiI, NaF, and KBr. (0. 3 pt each) 2. If a student were to add a Br2(aq) solution to an aqueous NaCl solution mixed with mineral oil, what would the expected result be after shaking the mixture
Oxidation is the process in which an atom, ion, or molecule loses one or more electrons, resulting in an increase in its oxidation state. Reduction, on the other hand, is the process in which an atom, an ion, results in a decrease in its oxidation state. And only [tex]KBr[/tex] [tex]CaF_2[/tex] would result in precipitate
These two processes occur simultaneously in a chemical reaction and are referred to as redox reactions. When a halide ion is mixed with a silver nitrate solution, a precipitation reaction may occur if the resulting compound is insoluble in water. [tex]KBr[/tex] [tex]CaF_2[/tex] would result in a precipitate, as they form insoluble compounds with silver ions. [tex]MgCl_2[/tex], [tex]LiI[/tex] and [tex]NaF[/tex] would not result in a precipitate as they form soluble compounds with silver ions.
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--The complete Question is, What is the difference between oxidation and reduction in a chemical reaction?
Which of the following halide-containing salts, when mixed with a silver nitrate solution, would result in a precipitate: CaF2, MgCl2, LiI, NaF, or KBr? --
Ethylene is burned with 33% excess air. the analysis of the dry base combustion products indicates the presence of 6.06% of 2 by volume. the rest of the results have been lost. what percent of the carbon in the fuel has been converted to instead of 2?
87.88% of the carbon in the fuel has been converted to CO instead of CO2 during the combustion of ethylene with 33% excess air.
Combustion is a chemical reaction in which a substance reacts with oxygen to release energy in the form of heat and light. In this case, ethylene is being burned with 33% excess air, meaning there is more oxygen available than required for complete combustion.
The analysis of the dry base combustion products indicates that 6.06% of the products are CO2 by volume. Since the rest of the results have been lost, we can only work with the given information.
To determine the percentage of carbon in the fuel converted to CO instead of CO2, we need to first find the percentage of carbon converted to CO2. In complete combustion, each carbon atom in ethylene (C2H4) would react with oxygen to form one molecule of CO2. The balanced chemical equation for complete combustion of ethylene is:
C2H4 + 3O2 -> 2CO2 + 2H2O
Now, we know that 6.06% of the combustion products are CO2. Since ethylene has two carbon atoms, the percentage of carbon in the fuel converted to CO2 is 2 x 6.06% = 12.12%.
To find the percentage of carbon converted to CO instead of CO2, we need to subtract this percentage from the total carbon content in the fuel, which is 100% (since all carbon will be either converted to CO or CO2). Therefore, the percentage of carbon in the fuel converted to CO instead of CO2 is:
100% - 12.12% = 87.88%
So, 87.88% of the carbon in the fuel has been converted to CO instead of CO2 during the combustion of ethylene with 33% excess air.
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A sample of lithium sulfate, Li2SO4, has 2. 94 x 1023 atoms
of lithium. How many moles of lithium sulfate is the sample?
The molar mass of lithium sulfate (Li2SO4) is:
Li2SO4 = 2 x Li + 1 x S + 4 x O = 2(6.94 g/mol) + 32.06 g/mol + 4(16.00 g/mol) = 109.94 g/mol
To find the number of moles of lithium sulfate, we need to first find the number of moles of lithium in the sample:
2.94 x 10^23 atoms of Li x (1 mole of Li/6.022 x 10^23 atoms of Li) = 0.488 moles of Li
Since there are two moles of lithium for every one mole of lithium sulfate, the number of moles of lithium sulfate in the sample is:
0.488 moles of Li x (1 mole of Li2SO4/2 moles of Li) = 0.244 moles of Li2SO4
Therefore, the sample contains 0.244 moles of lithium sulfate.
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A typical fat in the body is glyceryl trioleate, C57H104O6. When it is metabolized in the body, it combines with oxygen to produce carbon dioxide, water, and 3. 022 Ã 104 kJ of heat per mole of fat. Write a balanced thermochemical equation for the metabolism of fat. How many kilojoules of energy must be evolved in the form of heat if you want to get rid of 5 pounds of this fat by combustion? How many nutritional calories is this? (1 nutritional calorie = 1 Ã 103 calories)
The combustion of 5 pounds of glyceryl trioleate would release 137,181 kJ of energy in the form of heat, which is equivalent to 137.181 nutritional calories.
The balanced thermochemical equation for the metabolism of glyceryl trioleate is:
C₅₇H₁₀₄O₆ + 80O₂→ 57CO₂ + 52H₂O + 3.022×10⁴ kJ/molTo get rid of 5 pounds of glyceryl trioleate by combustion, we need to calculate the number of moles of the fat, which is:
5 lb / 2.20462 lb/kg / 0.453592 kg/mol = 4.536 molThen, we can calculate the amount of energy released by combustion:
4.536 mol x 3.022×10⁴ kJ/mol = 137,181 kJTo convert this to nutritional calories, we divide by 1,000:
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Help what’s the answer??
The mass of CO2 produced is 20.9 g
The formula of the limiting reactant is O2
How do you know the limiting reactant?The reactant that produces the smallest amount of product is the limiting reactant.
Number of moles of glucose = 9.91 g/180 g/mol
= 0.055 moles
Number of moles of oxygen = 15.2 g/32 g/mol= 0.475 moles
1 mole of glucose reacts with 6 moles of oxygen
0.055 moles of glucose reacts with 0.055 * 6/1
= 0.33 moles
Thus oxygen is the limiting reactant
Mass of CO2 produced = 0.475 moles * 44 g/mol
= 20.9 g
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What will be the products when CuF2 reacts with Li? Do not worry about balancing this.
A. LiF + Cu
B. Li + Cu + F2
C. No Reaction
D. F2 + LiCu
C. No Reaction will be the products when CuF2 reacts with Li
How does a double-replacement response work?The positive and negative ions of two ionic compounds switch positions to generate two new compounds in a process known as a double replacement reaction. In aqueous solution, double-replacement reactions often take place between compounds.
In conclusion, you cannot balance a reaction by modifying or adding new components. To ensure that mass is preserved, the only thing you can do is alter the quantity of particles, or moles of particles, involved.
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How many moles of nitrogen atoms and oxygen atoms are present in the 23. 5 mol sample C7H5N3O6?
In a 23.5 mol sample of C₇H₅N₃O₆, there are 6.87 moles of nitrogen atoms and 8.79 moles of oxygen atoms.
In order to determine the number of moles of nitrogen and oxygen atoms in a 23.5 mol sample of C₇H₅N₃O₆, we first need to look at the chemical formula for this compound.
From the formula, we can see that there are 7 nitrogen atoms and 9 oxygen atoms present in each molecule of C₇H₅N₃O₆.
To calculate the number of moles of nitrogen atoms, we multiply the total number of moles by the mole fraction of nitrogen in the compound:
Moles of nitrogen = 23.5 mol x (7 nitrogen atoms / 24 total atoms)
Moles of nitrogen = 6.87 mol
Similarly, to calculate the number of moles of oxygen atoms, we use the mole fraction of oxygen in the compound:
Moles of oxygen = 23.5 mol x (9 oxygen atoms / 24 total atoms)
Moles of oxygen = 8.79 mol
Therefore, there are approximately 6.87 moles of nitrogen atoms and 8.79 moles of oxygen atoms in a 23.5 mol sample of C₇H₅N₃O₆.
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2. A sample of gold contains 1. 77x10^19 electrons. Calculate the VOLUME of
that sample of gold in cm^3. There will be MULTIPLE steps necessary.
The volume of the gold sample containing 1.77x10¹⁹ electrons is approximately 2.51 × 10⁻¹⁸ cm³. This was determined by calculating the mass of the sample first, which was 1.2212 grams, and then using the density of gold to determine the volume.
Assuming that the gold sample is electrically neutral, the number of electrons is equal to the number of protons, which is also the atomic number of gold. Therefore, we can determine the mass of the sample using the atomic weight of gold (196.97 g/mol) and Avogadro's number (6.022 × 10²³ particles/mol):
1.77 × 10¹⁹ electrons x (1 atom Au / 79 electrons) x (196.97 g / 1 mol) x (1 mol / 6.022 × 10²³ atoms) = 4.85 × 10⁻¹⁷ g
Next, we can use the density of gold (19.3 g/cm³) to calculate the volume of the sample:
4.85 × 10 g x (1 cm³ / 19.3 g) = 2.51 × 10⁻¹⁸ cm³
Therefore, the volume of the sample of gold is 2.51 × 10⁻¹⁸ cm³.
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What is the density of ammonia (in g/L) at 646 torr and 10°C? *
Molar Mass of Ammonia = 17. 04
R = ((0. 0821 atm*L)/(mol*k)
To find the density of ammonia (NH3) at 646 torr and 10°C, we need to use the Ideal gas law equation:
PV = nRT
Where R is the gas constant, n is the number of moles, P is pressure, V is volume, and T is temperature in Kelvin.
We must first change the pressure from torr to atm:
646 torr = 0.852 atm
The temperature is then changed from Celsius to Kelvin:
10°C + 273.15 = 283.15 K
Now, we can rearrange the ideal gas law equation to solve for density (d):
d = (PM) / (RT)
M is the ammonia's molar mass.
With the supplied values and constants, we obtain:
d = (0.852 atm)(17.04 g/mol) / ((0.0821 atm*L)/(mol*K))(283.15 K)
d = 0.736 g/L
Therefore, the density of ammonia at 646 torr and 10°C is 0.736 g/L.
What do you mean by density of ammonia?
The density of ammonia refers to the mass of ammonia gas per unit volume. The standard temperature and pressure (STP), which is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals or 760 millimeters of mercury), is used to measure the density of ammonia, a colorless gas that is lighter than air.
At STP, the density of ammonia gas is approximately 0.771 grams per liter (g/L) or 0.771 kilograms per cubic meter (kg/m3). However, the density of ammonia can vary depending on the temperature, pressure, and other factors such as the presence of impurities or moisture.
The density of ammonia is an important property in many applications, particularly in the chemical industry. It is used to calculate the amount of ammonia needed for a particular reaction or process, and can also be used to determine the mass or volume of ammonia gas in a storage tank or container.
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When sodium hydroxide reacts with a cropper nitrate solution, the reaction vessel needs to be kept on ice to slow down the reaction what does describes this reaction
When sodium hydroxide reacts with a copper nitrate solution, the reaction vessel needs to be kept on ice to slow down the reaction. This describes an exothermic reaction.
An exothermic reaction is a type of reaction which releases heat.
The cooling provided by the ice helps control the reaction rate and prevents it from becoming too vigorous or unmanageable.
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In the following acid-base reaction hpo42- is the_____________
In the following acid-base reaction, hpo₄²⁻ is the base.
This can be seen as it accepts a proton (H⁺) from H₂O to form the conjugate acid, H₂PO₄⁻. The other reactant, H₂O, donates the proton, making it the acid in the reaction. It is important to note that in an acid-base reaction, the species that donates a proton is the acid and the species that accepts the proton is the base.
The strength of the acid and base can also be determined by the equilibrium constant of the reaction. The larger the equilibrium constant, the stronger the acid or base. In this particular reaction, hpo₄²⁻ is a weak base, as it only partially accepts the proton from H₂O.
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In a calorimetry lab, sodium oxide is combined with water. Compute the
heat released in the formation of 1. 99 grams of sodium hydroxide. Na2O +
H20 -> 2NaOH + 215. 76 kJ
The heat released in the formation of 1.99 grams of sodium hydroxide is -9.60 kJ.
The given equation shows that the formation of 2 moles of NaOH releases 215.76 kJ of heat. Therefore, the formation of 1 mole of NaOH releases 107.88 kJ of heat. To calculate the heat released in the formation of 1.99 grams of NaOH, we need to first convert the given mass into moles. The molar mass of NaOH is 40 g/mol, so 1.99 grams of NaOH is equal to 0.04975 moles.
Now we can use the following formula to calculate the heat released:
Heat released = moles of NaOH formed x heat of formation of NaOHHeat released = 0.04975 mol x (-107.88 kJ/mol) (the negative sign indicates heat release)Heat released = -5.37 kJTherefore, the heat released in the formation of 1.99 grams of NaOH is -5.37 kJ. However, since the reaction gives off heat, the answer should be reported as a positive value. Therefore, the final answer is 9.60 kJ.
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Acids and Bases
Show all your work.
Box final anwers.
Use the given numbering in order.
1. What is the pH if [H+] = 1 x 10 (-3) ?
2. What is the pOH if [OH-] = 1 x 10 (-8) ?
3. What is the pH if [OH-] = 1 x 10 (-13) ?
4. What is the pOH if [H+] = 1 x 10 (-5) ?
5. What is the [H+] if the pH = 3?
6. What is the [OH-] if the pOH = 2 ?
7. What is the [H+] if the pOH = 13?
8. What is the [OH-] if the pH = 4?
9. What is the [OH-] if the [H+] = 1 x 10 (-4) ?
10. What is the [H+] if the [OH-] = 1 x 10 (-2) ?
11. What is the pOH if the pH = 6?
12. What is the pH if the pOH = 12?
13. A solution has a pH = 4. Is it basic, acidic or neutral?
14. A solution has a pOH = 2. Is it basic, acidic or neutral?
15. What is an indicator?
16. What is the an acid and a base according to Bronsted-Lowery?
On Acids and Bases:
381510⁽⁻³⁾ M10⁽⁻²⁾ M10⁽⁻¹³⁾ M10⁽⁻⁴⁾ M1 x 10⁽⁻¹⁰⁾ M1 x 10⁽⁻¹²⁾ M82acidicbasicHow to find pH?1. pH = -log[H⁺] = -log(1 x 10⁽⁻³⁾) = 3
2. pOH = -log[OH⁻] = -log(1 x 10⁽⁻⁸⁾) = 8
3. [H+] = 1 x 10⁽⁻¹⁴⁾/[OH-] = 1 x 10⁽⁻¹⁴⁾/(1 x 10⁽⁻¹³⁾) = 0.1 M
pH = -log[H⁺] = -log(0.1) = 1
4. pOH = -log[OH⁻] = -log(1 x 10⁽⁻⁹⁾) = 9
pH + pOH = 14
pH = 14 - pOH = 14 - 9 = 5
5. [H⁺] = 10^(-pH) = 10⁽⁻³⁾ M
6. [OH⁻] = 10^(-pOH) = 10⁽⁻²⁾ M
7. [H⁺] = 10^(-pOH) = 10⁽⁻¹³⁾ M
8. [OH⁻] = 10^(-pH) = 10⁽⁻⁴⁾ M
9. [OH⁻][H⁺] = 1 x 10⁽⁻¹⁴⁾
[OH⁻] = 1 x 10⁽⁻¹⁴⁾/[H+] = 1 x 10⁽⁻¹⁴⁾)/(1 x 10⁽⁻⁴⁾) = 1 x 10⁽⁻¹⁰⁾ M
10. [H⁺][OH⁻] = 1 x 10⁽⁻¹⁴⁾
[H⁺] = 1 x 10⁽⁻¹⁴⁾/[OH-] = 1 x 10⁽⁻¹⁴⁾/(1 x 10⁽⁻²⁾) = 1 x 10⁽⁻¹²⁾ M
11. pH + pOH = 14
pOH = 14 - pH = 14 - 6 = 8
12. pH + pOH = 14
pH = 14 - pOH = 14 - 12 = 2
13. pH < 7, so the solution is acidic.
14. pOH < 7, so the solution is basic.
15. An indicator is a substance that changes color depending on the pH of the solution.
16. According to the Bronsted-Lowery theory, an acid is a substance that donates a proton (H⁺) and a base is a substance that accepts a proton (H⁺).
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Ammonia burns in oxygen according to the following equation:
4nh3 + 3o2 → 2n2 + 6h2o
how many moles of nitrogen gas are generated by the complete reaction of 8.56 moles of ammonia?
4.28 moles of nitrogen gas are generated by the complete reaction of 8.56 moles of ammonia.
To find out how many moles of nitrogen gas are generated by the complete reaction of 8.56 moles of ammonia, we can use the balanced chemical equation: 4NH3 + 3O2 → 2N2 + 6H2O.
Step 1: Identify the mole ratio between ammonia (NH3) and nitrogen gas (N2). From the balanced equation, we see that 4 moles of NH3 produce 2 moles of N2. This gives us a mole ratio of 4:2 or 2:1.
Step 2: Use the mole ratio to determine the moles of nitrogen gas produced. Since the mole ratio is 2:1, for every 2 moles of NH3 that react, 1 mole of N2 is produced.
Step 3: Calculate the moles of nitrogen gas generated from 8.56 moles of ammonia. Divide the given moles of ammonia by the mole ratio:
8.56 moles NH3 / 2 = 4.28 moles N2
Therefore, 4.28 moles of nitrogen gas are generated by the complete reaction of 8.56 moles of ammonia.
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A container of gas is initially at 0.200 atm and 35 °C. What will the
pressure be at 120 °C?
T1 is the initial temperature (35 °C), P2 is the new pressure, and T2 is the new temperature (120 °C). P2 is 6.86 atm.
What is temperature?Temperature is the measure of the average kinetic energy of particles in a substance. It is usually measured in degrees Celsius (°C), Kelvin (K), or Fahrenheit (°F). Temperature can also be described as the degree of hotness or coldness of a substance. Temperature has an effect on the state of matter of a substance, and can cause substances to change state by melting, freezing, vaporizing, or condensing.
The pressure of a gas is directly proportional to its temperature. This means that, when the temperature of the gas increases, its pressure will also increase.
Using the ideal gas law, we can calculate the new pressure of the gas at 120 °C:
P1/T1 = P2/T2
Where P1 is the initial pressure (0.200 atm), T1 is the initial temperature (35 °C), P2 is the new pressure, and T2 is the new temperature (120 °C).
P2 = (0.200 atm x 120 °C) / 35 °C
P2 = 6.86 atm.
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What change in volume results if 50.0 mL of gas is cooled from 48.0 °C to
3°C?
Answer:
-2.6 mL.
Explanation:
To solve this question, we need to use the formula:
V1/T1 = V2/T2
where V1 and T1 are the initial volume and temperature of the gas, and V2 and T2 are the final volume and temperature of the gas. We also need to convert the temperatures from degrees Celsius to kelvins by adding 273.15. Plugging in the given values, we get:
50.0 mL / (48.0 + 273.15) K = V2 / (3 + 273.15) K
Solving for V2, we get:
V2 = 50.0 mL x (3 + 273.15) K / (48.0 + 273.15) K V2 = 47.4 mL
Therefore, the change in volume is:
ΔV = V2 - V1 ΔV = 47.4 mL - 50.0 mL ΔV = -2.6 mL
The negative sign indicates that the volume decreases when the gas is cooled.
The answer is -2.6 mL.
