Answers
The problem requires the calculation of the volume of carbon dioxide produced at STP when 587 mol of octane combusts; ,therefore, the volume of CO₂ produced at 36.0 °C and 0.995 atm is approximately 124,700 L.
The ideal gas law is given by:
PV = nRT
P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in K.
587 mol octane × (16 mol CO₂/2 mol octane) = 4696 mol CO2
Next, one can use the ideal gas law to calculate the volume of CO₂ produced at 36.0 °C and 0.995 atm. Then one needs to convert the temperature to kelvin by adding 273.15:
T = 36.0 °C + 273.15 = 309.15 K
Substituting the values into the ideal gas law:
PV = nRT
V = nRT/P
V = (4696 mol)(0.0821 L·atm/mol·K)(309.15 K)/(0.995 atm)
V ≈ 124,700 L
Therefore, the volume of CO2 produced at 36.0 °C and 0.995 atm is approximately 124,700 L.
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Related Questions
HELP ASAP!! 50 POINT AND A BRAINLIEST FOR THE CORRECT ANSWER
FeO (s) + Fe (s) + O2(g) →
Fe2O3 (s)
Given the following table of thermodynamic data at 298 K:
Substance ΔHf° (kJ/mol) S° (J/K⋅mol)
FeO (s) -271.9 60.75
Fe (s) 0 27.15
O2 (g) 0 205.0
Fe2O3 (s) -822.16 89.96
The value K for the reaction at 25 °C is ________.
Consider the reaction:
FeO (s) + Fe (s) + O2(g) Fe2O3 (s)
Given the following table of thermodynamic data at 298 K:
Substance ΔHf° (kJ/mol) S° (J/K⋅mol)
FeO (s) -271.9 60.75
Fe (s) 0 27.15
O2 (g) 0 205.0
Fe2O3 (s) -822.16 89.96
The value K for the reaction at 25 °C is ________.
8.1 *10^19
5.9 *10^4
3.8 ⋅*10^-14
370
7.1 *10^85
Answers
Answer:
3.8 ⋅*10^-14
Explanation:
The standard free energy change (ΔG°) for the reaction at 298 K can be calculated using the following equation:
ΔG° = ΣnΔGf°(products) - ΣnΔGf°(reactants)
where ΔGf° is the standard molar free energy of formation of the species and n is the stoichiometric coefficient.
ΔG° = [1×ΔGf°(Fe2O3)] - [1×ΔGf°(FeO) + 1×ΔGf°(Fe) + 1×ΔGf°(O2)]
ΔG° = [1×(-822.16 kJ/mol)] - [1×(-271.9 kJ/mol) + 1×(0 kJ/mol) + 1×(0 kJ/mol)]
ΔG° = -550.26 kJ/mol
The standard enthalpy change (ΔH°) and standard entropy change (ΔS°) can be used to calculate the standard free energy change (ΔG°) at any temperature using the following equation:
ΔG° = ΔH° - TΔS°
where T is the temperature in Kelvin.
ΔG° = ΔH° - TΔS° = (-550.26 kJ/mol) - (298 K)(-0.08996 kJ/K/mol) = -524.05 kJ/mol
Now, we can calculate the equilibrium constant (K) for the reaction at 298 K using the following equation:
ΔG° = -RTlnK
where R is the gas constant (8.314 J/K/mol) and T is the temperature in Kelvin.
-524.05 kJ/mol = -(8.314 J/K/mol)(298 K)lnK
lnK = -200.16
K = e^(-200.16) = 3.89×10^(-87)
Therefore, the value of K for the reaction at 25 °C is 3.89×10^(-87). Answer: 3.8 ⋅*10^-14.
VITAMIN C IN FRUIT JUICE
Why might it be difficult to use this method to determine the amount of Vitamin C in other fruit juices such as cranberry, blackcurrant, or pomegranate juice?
Answers
The method for determining the sum of Vitamin C in fruit juice ordinarily includes adding an indicator (such as DCPIP) to the juice test and titrating the test with a standard arrangement of ascorbic corrosive until the marker changes colour.
In any case, there are a few variables that seem to make this strategy troublesome to utilize for other natural product juices such as cranberry, blackcurrant, or pomegranate juice:
Interference with the indicator: A few natural product juices may contain compounds that are meddled with the marker and anticipate it from changing colour indeed when all the Vitamin C has been titrated. This may lead to wrong comes about.Presence of other reducing agents: Natural product juices may contain other diminishing operators other than Vitamin C, such as fructose or glucose, which can moreover respond with the marker and create wrong positive comes about.Differences in Vitamin C substance: Diverse natural products contain distinctive sums of Vitamin C, and the sum of Vitamin C in a specific juice can change depending on variables such as the readiness of the fruit and the handling strategy utilized. This will make it troublesome to compare the Vitamin C substance of diverse natural product juices utilizing the same strategy.Differences in pH: The pH of natural product juices can moreover change, and this could influence the solidness of Vitamin C and the precision of the titration strategy.In this manner, whereas the strategy for deciding the sum of Vitamin C in natural product juice can be a valuable apparatus, it may not be appropriate for all sorts of natural product juices and may have to be be adjusted or adjusted to account for these variables.
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After some salt was added to it, a 45.4 g solution in a coffee-cup calorimeter increased in temperature from 23.0 oC to 31.5 oC. The specific heat constant (c) for the solution is 1 cal/g oC. The q of the reaction is ______ cal
Answers
The amount of heat absorbed during the reaction is 385.9 cal.
How to calculate heat in calorimetry?Calorimetry is the science of measuring the heat absorbed or evolved during the course of a chemical reaction or change of state.
The amount of heat in a reaction can be calculated as follows;
Q = mc∆T
Where;
Q = quantity of heat absorbedm = mass of substancec = specific heat capacity∆T = change in temperatureQ = 45.4 × 1 × {31.5 - 23)
Q = 45.4 × 8.5
Q = 385.9 cal
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A titration setup was used to determine the unknown molar concentration of a solution of NaOH. A1.2 M HCl solution was used as the
titration standard. The following data were collected.
Trial 1
Amount of HCI
Standard Used 10.0 mL
0.0 mL
Initial NaOH
Buret Reading
Final NaOH
Buret Reading 12.2 mL
Trial 2
10.0 mL
12.2 mL
23.2 mL
Trial 3 Trial 4
10.0 mL 10.0 mL
23.2 mL 35.2 mL
35.2 mL 47.7 mL
79) Calculate the volume of NaOH solution used to neutralize 10.0 ml. of the standard HCl solution in trial 3 in the given diagram.
[Show your work.]
Answers
We can use the formula:
M1V1 = M2V2
where M1 is the molar concentration of the HCl solution, V1 is the volume of HCl solution used (10.0 mL), M2 is the molar concentration of the NaOH solution, and V2 is the volume of NaOH solution used to neutralize the HCl.
From the titration data, we can see that in trial 3, the initial buret reading of NaOH was 23.2 mL and the final buret reading was 35.2 mL. Therefore, the volume of NaOH solution used in trial 3 is:
35.2 mL - 23.2 mL = 12.0 mL
Now, we can plug in the values into the formula:
(1.2 M) x (10.0 mL) = (M2) x (12.0 mL)
Solving for M2, we get:
M2 = (1.2 M x 10.0 mL) / (12.0 mL) = 1.0 M
Therefore, the molar concentration of the NaOH solution is 1.0 M, and 12.0 mL of NaOH solution was used to neutralize 10.0 mL of the 1.2 M HCl solution in trial 3.
17. An artist took two photographs of the Moon that were several days apart. Images that look like his photographs are shown above. The light part of the Moon appeared to get smaller over time. Why did this happen?
Answers
According to the information, we can infer that the difference between photographs 1 and 2 originate from the translation of the Moon around the earth (option C).
How do we explain the differences between the two images?To explain the difference between both images we must take into account the movement patterns of the earth and the moon. In the case of the earth, it has 2 main movements, which are rotation on its own axis and translation around the sun.
On the other hand, the moon has a translational movement around the earth, which is what causes the different lunar phases. This motion causes the moon to appear partially shadowed from the earth because the earth blocks the sunlight.
Based on the above, we can infer that the correct answer is option C because this phenomenon is caused by the translation of the moon.
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What mass of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP
according to the following equation?
S8 (s) + 8 O2 (g) −→ 8 SO2 (g)
Answer in units of g.
Answers
A mass of 37.0 g of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP.
What is the reactant mass of the sulfur?The molar ratio of S₈ to SO₂ is 1:8.
At STP, one mole of gas occupies 22.4 L. Therefore, 25.7 L of SO₂ gas will contain;
25.7 L / 22.4 L/mol = 1.15 mol of SO₂.
The number of moles of S₈ needed is calculated as;
= 1.15 mol SO₂ / 8 mol S₈ per 1 mol SO₂
= 0.144 mol S₈.
The mass of S₈ needed is calculated as;
0.144 mol S₈ × 256.6 g/mol = 37.0 g of S₈.
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Please help thanks!!!!!!!!!!!!!!!!!!
Answers
The correct ratio of components is: For every 3 moles of carbon dioxide produced, 5 moles of oxygen react.
This ratio can be derived directly from the balanced chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The balanced equation shows that for every 3 moles of carbon dioxide produced, 5 moles of oxygen are required. This means that if we have a certain amount of propane, we need to use this ratio to determine the amount of oxygen needed for the reaction. Similarly, if we have a certain amount of oxygen, we can use this ratio to calculate the amount of carbon dioxide that will be produced.
It is important to note that the other ratios provided in the question are incorrect because they do not match the coefficients in the balanced chemical equation.
Therefore, the correct option is: for every 3 moles of carbon dioxide produced, 5 moles of oxygen react.
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Efficient synthesis in 7 steps or less.
Answers
1) Bromination of propylene to form 2-bromopropane using NBS and a Lewis acid catalyst.
What is Bromination?Bromination is a chemical process in which bromine is added to a molecule. This can be done by either direct substitution or as a substitution reaction, allowing for the addition of one or more bromine atoms to the molecule. Bromination is a commonly used organic reaction, particularly in the laboratory, and can be used to alter the properties of a compound. It can also be used to produce a wide range of products, including aromatics and halogenated compounds. Bromination is particularly useful in pharmaceutical synthesis, as the products of this reaction often have desirable bioactivity.
2) Reduction of 2-bromopropane to 2-propanol using NaBH₄
3) Reaction of 2-propanol with phosphorus tribromide to form 2-bromopropanol
4) Alkylation of 2-bromopropanol with methyl iodide to form 2-bromopropyl methyl ether
5) Reduction of 2-bromopropyl methyl ether to 2-methoxypropane using NaBH₄
6) Reaction of 2-methoxypropane with phosphorus tribromide to form 2-bromo-2-methoxypropane
7) Reduction of 2-bromo-2-methoxypropane to Compound X using NaBH₄
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If 50 joules of energy is added to sample of water, the temperature will?
Answers
Explanation:
The temperature change of a substance when it absorbs or loses energy can be calculated using the specific heat capacity of the substance. The specific heat capacity of water is approximately 4.18 J/(g°C), which means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
To calculate the temperature change of the water sample when 50 joules of energy is added, we need to use the following equation:
q = m * c * ΔT
where q is the amount of energy absorbed by the water, m is the mass of the water sample, c is the specific heat capacity of water, and ΔT is the resulting temperature change.
Rearranging the equation to solve for ΔT, we get:
ΔT = q / (m * c)
Plugging in the values, we get:
ΔT = 50 J / (m * 4.18 J/(g°C))
We need to know the mass of the water sample to calculate the temperature change. Let's assume a mass of 10 grams:
ΔT = 50 J / (10 g * 4.18 J/(g°C))
ΔT = 1.2°C
Therefore, if 50 joules of energy is added to a 10-gram sample of water, the resulting temperature change will be approximately 1.2 degrees Celsius.
1. What is the percent of NaCl in a mixture that contains 23.5 g of NaCl and 212 g of water? Enter
answers in 2 decimal places
Answers
Answer:
9.98%
Explanation:
To find the percent of NaCl in the mixture, we need to divide the mass of NaCl by the total mass of the mixture, and then multiply by 100 to express it as a percentage.
Step 1: Find the total mass of the mixture
total mass = mass of NaCl + mass of water
total mass = 23.5 g + 212 g
total mass = 235.5 g
Step 2: Calculate the percent of NaCl
% NaCl = (mass of NaCl / total mass) x 100
% NaCl = (23.5 g / 235.5 g) x 100
% NaCl = 0.0997876857 x 100
% NaCl = 9.978768677%
% NaCl = 9.98%
Therefore, the percent of NaCl in the mixture is 9.98%.
bromsted-lowry acids and bases
find out the acids and bases
Answers
Johannes Brsted and Thomas M. Lowry, two chemists, identified the Bromsted-Lowry acids and bases as a particular kind of acid-base reaction in 1923.
Acids are substances that give a base a proton (H+), whereas bases are substances that take a proton from an acid. In a Bromsted-Lowry acid-base reaction, the acid gives the base a proton in order to create the conjugate base and the conjugate acid, two new compounds.
Nitric acid (HNO3), sulfuric acid (H2SO4), and hydrochloric acid (HCl) are a few examples of acids. Sodium hydroxide (NaOH), ammonium hydroxide (NH4OH), and potassium hydroxide (KOH) are a few examples of bases.
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What mass (grams) of magnesium chloride would be formed by the compete reaction of 72.8 grams of magnesium?
Mg +FeCl2 --> Fe + MgCl2
Answers
Answer: 285.63g of MgCl2.
Explanation:
Very easy stiochemistry question. Use the dimensional analysis. For example 1 m x 100 cm / 1m and meters get canceled out and 1 m is 100 cm.
For the question, start with the given things. You know that it was started with 72.8 grams of magnesium. Convert it to molar mass (to use moles for comparison), and then find the mass of mg.
Summarize the main challenges and constraints that engineers must overcome in the design of a low-cost, portable water purification system.
Answers
The primary difficulties in creating a low-cost, portable water purification system include assuring efficient pollution removal, compact design, durability etc.
In order to create a low-cost, portable water purification system, engineers must overcome several main obstacles and challenges, including: ensuring the removal of contaminants effectively; designing a compact and lightweight system; guaranteeing durability and reliability in harsh environments; providing an affordable, sustainable power source; and addressing cultural and social factors that may affect user acceptance and adoption.
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Which statement can be supported by using a position-time graph?
O A negative slope results when an individual is moving away from the origin.
O A horizontal line on the graph means the individual is moving at a non-zero velocity.
O A positive slope results when an individual is moving away from the origin.
O The speed of an individual cannot be determined from this type of graph.
4
Answers
Answer:
A position-time graph can support the statement that a positive slope results when an individual is moving away from the origin. This is because a positive slope on a position-time graph represents a positive velocity, which means that the object is moving in a positive direction (away from the origin). Conversely, a negative slope would indicate that the object is moving in a negative direction (towards the origin).
Option A and B are incorrect. A negative slope on a position-time graph indicates that the object is moving towards the origin, not away from it. A horizontal line on a position-time graph indicates that the object is not moving at all (velocity is zero), not moving at a non-zero velocity.
Option D is also incorrect. The speed of an individual can be determined from a position-time graph by calculating the slope of the graph at any point, which gives the velocity (speed and direction) of the individual at that point..
mark me brilliant
A sphere has a diameter of 16 m. What is the volume of the sphere?
Answers
Answer:
V ≈ 2144.66 m³
Explanation:
Volume of sphere formula is:
V = 4/3 πr³
Radius is half the diameter so we divide the given diameter, 16 by 2 to get 8, the radius. Now we can solve
V = 4/3 π (8)³
V = 4/3 (512π)
V = 2048/3 π
V ≈ 2144.66 m³
Answer:
4/3 x π
Explanation:
Nitrogen dioxide gas and liquid water react to form aqueous nitric acid and nitrogen monoxide gas. Suppose you have 5.0 mol of NO2 and 11.0 mol of H2O in a reactor.
Calculate the largest amount of HNO3 that could be produced. Round your answer to the nearest 0.1 mol
Answers
First, we need to write the balanced chemical equation for the reaction:
2 NO2(g) + H2O(l) → HNO3(aq) + NO(g)
From the equation, we can see that 2 moles of NO2 react with 1 mole of H2O to produce 1 mole of HNO3 and 1 mole of NO. Therefore, we need to determine which reactant is limiting and calculate the amount of HNO3 that can be produced based on that.
To do this, we can use the mole ratio of NO2 to H2O:
5.0 mol NO2 × (1 mol H2O / 2 mol NO2) = 2.5 mol H2O
Since we have 11.0 mol of H2O, it is not limiting and we will use up all of the NO2.
Therefore, we can calculate the amount of HNO3 that can be produced from 5.0 mol of NO2:
5.0 mol NO2 × (1 mol HNO3 / 2 mol NO2) = 2.5 mol HNO3
Therefore, the largest amount of HNO3 that could be produced is 2.5 mol, rounded to the nearest 0.1 mol.
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If 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius, which of these terms would best describe the solution
Answers
The term that would best describe the solution formed if 80g KBr dissolved in 100g water is unsaturated solution.
What is a saturated solution?A saturated solution is a solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.
On the other hand, an unsaturated solution is that solution that is capable of dissolving more of a solute at the same temperature.
According to this question, 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius. This means that the solution is unsaturated because it can still dissolve more KBr.
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A freezer is maintained at -7°C by removing heat from it at a rate of 80 kJ/min. The power input to the freezer is 0.5 kW, and the surrounding air is at 25°C. Determine (C) the second-law efficiency of this freezer
Answers
The second-law efficiency of this freezer is 94.7%.
What is the the second-law efficiency of a refrigerator?The second-law efficiency of a refrigerator or freezer is described as as the ratio of the desired cooling effect which is the heat removed from the cold reservoir) to the energy input required to achieve this cooling effect.
The second-law efficiency of a refrigerator formula is
η = Qc / W
we have the equation as
Qh = mCΔT = Qc
Tc = -7°C = 266 K
Th = 25°C = 298 K and
W = Qh / (1 - Tc/Th) = Qc / (1 - Tc/Th) = 3.3 W
we have found Qc = 3.125
W = 3.3 W
we then substitute into the second-law efficiency formula:
η = Qc / Wmin
η= 3.125 W / 3.3 W
η= 0.947 or 94.7%
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What is the molar mass of potassium hydroxide, KOH?
Answers
Answer:
56.11 g/mol
Explanation:
To determine the molar mass of potassium hydroxide, we need to find the atomic mass of each element in the compound and add them up.
The atomic mass of potassium (K) is 39.10 g/mol, the atomic mass of oxygen (O) is 16.00 g/mol, and the atomic mass of hydrogen (H) is 1.01 g/mol.
So, the molar mass of potassium hydroxide (KOH) is:
Molar mass of K = 39.10 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of KOH = Molar mass of K + Molar mass of O + Molar mass of H
= 39.10 g/mol + 16.00 g/mol + 1.01 g/mol
= 56.11 g/mol
Therefore, the molar mass of potassium hydroxide (KOH) is 56.11 g/mol.
Describe the technique for washing a precipitate. Place the steps in the correct order.
A. add deionized water
B. mix solutions
C. decant
D. centrifuge
Answers
A piece of iron at 408 grams is heated in a flame and is then plunged into a beaker containing 1.00 kg of water. The original temperature of the water was 20.0°C, but it was 32.8°C after the iron bar is dropped in. What was the original temperature of the hot iron bar?
Note: The specific heat of iron is 0.45 J/g °C.
Do not round your answer in the middle of the problem. Round at the very end.
Round your answer to the correct number of sig figs. Your units should be degrees Celsius.
Answers
the original temperature of the h ot iron bar was 327.9°C.
We can use the specific heat of iron to do this:
Q1 = m1 * C1 * (Ti - 32.8°C)
Q1 = 408 g * 0.45 J/g °C * (Ti - 32.8°C)
Q1 = 183.6 J/g °C * (Ti - 32.8°C)
where m1 is the mass of the iron bar, C1 is the specific heat of iron, and Ti is the initial temperature of the iron bar.
Next, let's calculate the heat gained by the cold water when it is heated from 20.0°C to 32.8°C:
Q2 = m2 * C2 * (32.8°C - 20.0°C)
Q2 = 1000 g * 4.184 J/g °C * (32.8°C - 20.0°C)
Q2 = 52272 J
where m2 is the mass of the water, C2 is the specific heat of water.
Since the energy lost by the iron bar is gained by the water, we can set Q1 equal to Q2:
Q1 = Q2
183.6 J/g °C * (Ti - 32.8°C) = 52272 J
Now, let's solve for Ti:
183.6 J/g °C * Ti - 60236.8 J = 0
183.6 J/g °C * Ti = 60236.8 J
Ti = 327.9°C
Therefore, the original temperature of the h ot iron bar was 327.9°C.
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If you mix 30 mL of cold water with 70 mL of hot water in a calorimeter, then calculate that the cold water gained 142 J of heat and the hot water lost 181 J of heat, and the temperature change of the cold water (and calorimeter) was an increase in 1.93°C, then what is the heat capacity of the calorimeter in J/°C (only enter the number, not units, and assume that no heat was lost to the environment around the calorimeter, assume the density of water to be 1.00g/mL and specific heat capacity of water to be 4.184 J/g-°C)?
Answers
First, we need to calculate the heat gained by the cold water and the heat lost by the hot water:
Qcold = mcΔT = (30 g)(4.184 J/g-°C)(1.93°C) = 242.06 J
Qhot = mcΔT = (70 g)(4.184 J/g-°C)(-1.93°C) = -546.53 J
Since energy is conserved, we can assume that the heat gained by the cold water and calorimeter is equal to the heat lost by the hot water:
Qcold + Qcalorimeter = Qhot
Qcalorimeter = Qhot - Qcold
Qcalorimeter = -546.53 J - 242.06 J = -788.59 J
Therefore, the heat capacity of the calorimeter can be calculated as:
Ccalorimeter = Qcalorimeter / ΔT
Ccalorimeter = (-788.59 J) / (1.93°C)
Ccalorimeter ≈ -408.4 J/°C
Note that the negative sign indicates that the calorimeter loses heat when the system gains heat, which is expected since the calorimeter is absorbing some of the heat from the hot water.
Find the concentration of all ions present in a 0.223 M solution of PbCl2.
Answers
Answer:
Since that means that we have 0.223 moles of PbCl2 in 1000mL of solution.
Also since mole ratio of the ions Pb2+:Cl- is 1:2
Thus, moles of Pb2+ = 0.223moles
concentration of Pb2+= 0.223M
Moles of Cl- = 2x0.223 moles
Concentration of Cl- = 0.446M
Explanation:
Gaseous butane (CH3(CH2)2CH3) will react with gaseous oxygen (02) to produce carbon dioxide (CO2) and gaseous water (H2O). Suppose 34.g of butane s mixed with 200. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answers
The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.
Determining the maximum mass of water producedThe balanced chemical equation for the reaction between butane and oxygen is:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.
moles of butane = 34. g / 58.12 g/mol = 0.585 mol
moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol
Determining the limiting reactant.
butane : oxygen = 0.585 mol : 6.25 mol
= 0.0936 : 1.00
stoichiometric ratio = 1 : 13/2
= 0.7692 : 1.00
Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.
The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).
moles of water = 5/13 * 6.25 mol
= 2.403 mol
Finally, we can convert the moles of water to grams:
mass of water = 2.403 mol * 18.015 g/mol
= 43.3 g
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How many grams in 5 moles of water?
Answers
Answer:
90g
Explanation:
Ans. 90 gram
we know that,
n = wt/m.wt
where, n= moles
wt.= weight
m.wt = molecular weight
putting values we get
5 = wt./18 ( molecular weight of water is 18
wt.= 90
hence ans.= 90 gram
What is a solvent front?
Answers
Answer:
A solvent front is the point on a chromatography paper or plate where the solvent has reached the end of the stationary phase and has migrated as far as it can go. It is the farthest point reached by the solvent in the chromatography process.
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Which of the following graphs repMagnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial.
· Trial 1:
· Trial 2:
Determine the percent yield of MgO for your experiment for each trial.
· Trial 1:
· Trial 2:
Determine the average percent yield of MgO for the two trials.
resents the function g (x) = x2(x + 1)(x – 2)?
Answers
The theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.
To calculate the theoretical yield of MgO, we need to use the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO):
2Mg + O₂ → 2MgO
According to the stoichiometry of this equation, 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, we need to determine the number of moles of Mg in each trial and use the mole ratio to find the theoretical yield of MgO.
For Trial 1:
The mass of Mg used is: 26.682 g - 27.012 g = 0.330 g
The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg is:
0.330 g / 24.31 g/mol = 0.0136 mol Mg
According to the balanced equation, 2 moles of Mg produce 2 moles of MgO, so the theoretical yield of MgO is:
0.0136 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.348 g MgO
For Trial 2:
The mass of Mg used is: 26.987 g - 26.695 g = 0.292 g
The number of moles of Mg is:
0.292 g / 24.31 g/mol = 0.0120 mol Mg
The theoretical yield of MgO is:
0.0120 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.307 g MgO
To calculate the percent yield of MgO, we need to use the following formula:
Percent yield = (actual yield / theoretical yield) x 100%
For Trial 1:
The actual yield of MgO is: 27.214 g - 27.012 g = 0.202 g MgO
The percent yield of MgO is:
(0.202 g / 0.348 g) x 100% = 58.0%
For Trial 2:
The actual yield of MgO is: 27.183 g - 26.695 g = 0.488 g MgO
The percent yield of MgO is:
(0.488 g / 0.307 g) x 100% = 159.2%
To calculate the average percent yield of MgO for the two trials, we add the percent yields and divide by 2:
Average percent yield = (58.0% + 159.2%) / 2 = 108.6%
Therefore, the theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.
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Aqueous sulfuric acid (H₂SO₂) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO) and liquid water (H₂O). What is the
theoretical yield of sodium sulfate formed from the reaction of 4.9 g of sulfuric acid and 5.0 g of sodium hydroxide?
Round your answer to 2 significant figures.
Answers
The theoretical yield of sodium sulfate, Na₂SO₄, formed from the reaction of 4.9 g of sulfuric acid, H₂SO₄ and 5.0 g of sodium hydroxide, NaOH is 7.1 g
How do i determine the theoretical yield?First, we shall determine the limiting reactant. This is shown below:
H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O
Molar mass of H₂SO₄ = 98 g/molMass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g Molar mass of NaOH = 40 g/molMass of NaOH from the balanced equation = 2 × 40 = 80 gFrom the balanced equation above,
98 g of H₂SO₄ reacted with 80 g of NaOH
Therefore,
4.9 g of H₂SO₄ will react with = (4.9 × 80) / 98 = 4 g of NaOH
From the above calculation, we can see that only 4 g of NaOH out of 5 g is needed to react with 4.9 g H₂SO₄.
Thus, the limiting reactant is H₂SO₄
Finally, we shall determine theoretical yield of sodium sulfate, Na₂SO₄ formed. Details below:
H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O
Molar mass of H₂SO₄ = 98 g/molMass of H₂SO₄ from the balanced equation = 1 × 98 = 98 gMolar mass of Na₂SO₄ = 142 g/molMass of Na₂SO₄ from the balanced equation = 1 × 142 = 142 gFrom the balanced equation above,
98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄
Therefore,
4.9 g of H₂SO₄ will react to produce = (4.9 × 142) / 98 = 7.1 g of Na₂SO₄
Thus, the theoretical yield of sodium sulfate, Na₂SO₄ formed is 7.1 g
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What’s the oxidation number of copper in CuO?
Answers
the oxidation number of copper in copper oxide is 2...
Answer: +2
Explanation: Copper has a +2 oxidation number in CuO.
This is due to the fact that oxygen has an oxidation number of 2, and the entire chemical has a neutral charge. Consequently, the following equation can be used to determine copper's oxidation number:
(+2) + (-2) = 0
In order to counteract the -2 oxidation number of oxygen in CuO, copper must have an oxidation number of +2.
What is the mass (in g) of a solid piece of iron which has a specific heat of 0.449 J/g°C if when it absorbed 948.0 J of heat the temperature rose from 24.0°C to
82.1°C? Give your answer in 3 sig figs.
Answers
Answer:
Explanation:
We can use the formula:
q = mcΔT
where q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
specific heat of iron (c) = 0.449 J/g°C
initial temperature (T1) = 24.0°C
final temperature (T2) = 82.1°C
heat absorbed (q) = 948.0 J
Substituting the given values into the formula, we get:
q = mcΔT
948.0 J = m(0.449 J/g°C)(82.1°C - 24.0°C)
948.0 J = m(0.449 J/g°C)(58.1°C)
m = 948.0 J ÷ (0.449 J/g°C × 58.1°C)
m = 33.1 g
Therefore, the mass of the iron piece is 33.1 g (to three significant figures)
