The Snickers bar released 1,077,280 joules of energy when burned.
To calculate the energy released by burning a Snickers bar, we can use the formula:
q = mcΔT
where q is the heat energy released, m is the mass of water, c is the specific heat of water, and ΔT is the temperature change.
We know the mass of water is 3500 g, and the temperature change is 72°C. The specific heat of water is 4.184 J/g°C.
Therefore:
q = (3500 g) x (4.184 J/g°C) x (72°C) = 1077280 J
So, the Snickers bar released 1,077,280 joules of energy when burned.
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Consider a gas cylinder containing 0. 100 moles of an ideal gas in a
volume of 1. 00 L with a pressure of 1. 00 atm. The cylinder is
surrounded by a constant temperature bath at 298. 0 K. With an
external pressure of 5. 00 atm, the cylinder is compressed to 0. 500 L.
Calculate the q(gas) in J for this compression process.
According to the question the q(gas) in J for this compression process is 0J.
What is gas ?Gas is a state of matter in which particles are spread out and have enough energy to move around freely. Gas is composed of molecules in constant motion and takes the shape and volume of its container. Gas can be either naturally occurring or man-made and is found in the atmosphere. Examples of naturally occurring gases include oxygen, nitrogen, and carbon dioxide. Man-made gases include helium, chlorine, and hydrogen. Gas is often used as a source of energy and is burned to produce heat, which can be used to power machines and vehicles. Gas is also used in many industries, such as in the production of chemicals and plastics.
In this case, n = 0.100 moles,
[tex]C_v[/tex] = (3/2)R = (3/2)(8.314 J/mol K) = 12.471 J/mol K, and
T₁ = 298.0 K,
T2 = 298.0 K.
Therefore, q(gas)
= nCv (T₂- T₁)
= 0.100 mol × 12.471 J/mol K × (298.0 K - 298.0 K)
= 0 J.
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a 90.-ml sample of juice was titrated with the i2(aq) solution described above using a buret. the initial reading of the buret was 0.24 ml. when the endpoint was reached, the reading on the buret was 33.08 ml. how many mg of vitamin c were in the juice sample?
The juice sample contains 28,920 mg of vitamin C.
The amount of iodine used in the reaction can be calculated as:
I2 used = (final buret reading - initial buret reading) * 0.005 M
I2 used = (33.08 ml - 0.24 ml) * 0.005 M = 0.16392 moles
Since 1 mole of vitamin C reacts with 1 mole of iodine, the amount of vitamin C in the juice can be calculated as:
Vitamin C = I2 used * (1 mol of vitamin C / 1 mol of I2) * (176.12 g/mol)
Vitamin C = 0.16392 * (1 / 1) * (176.12 g/mol) = 28.92 g
Converting to milligrams:
Vitamin C = 28.92 g * 1000 mg/g = 28,920 mg
Therefore, the juice sample contains 28,920 mg of vitamin C.
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1. in a laboratory experiment, an undergraduate student collected a sample of ammonium
phosphate. if the sample contains 9.52 x 1025 molecules, how many grams of the sample did he
collected?
The student collected 2.63 x 10¹⁰ grams of ammonium phosphate.
To determine the mass of the sample collected, we need to know the molar mass of ammonium phosphate, which is (NH₄)₃PO₄. The molar mass of (NH₄)₃PO₄ can be calculated by adding the atomic masses of the constituent atoms:
Molar mass of (NH₄)₃PO₄ = (3 x molar mass of NH₄) + (1 x molar mass of PO₄)
= (3 x 18.04 g/mol) + (1 x 94.97 g/mol)
= 149.99 g/mol
The number of moles of (NH₄)₃PO₄ in the sample can be calculated by dividing the number of molecules by Avogadro's number (6.022 x 10²³):
Number of moles of (NH₄)₃PO₄ = 9.52 x 10²⁵ molecules / 6.022 x 10²³ molecules/mol
= 15.8 mol
Finally, we can calculate the mass of the sample using the formula:
Mass = Number of moles x Molar mass
= 15.8 mol x 149.99 g/mol
= 2.63 x 10¹⁰ g
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Using an experimentally determined value (1. 8×10−10) of Ksp, determine the value for the reaction quotient 'Q' if Ag2CrO4 will precipitate when 5. 00 mL of 0. 0040 M AgNO3 are added to 4. 00 mL of 0. 0024 M K2CrO4
The solubility product constant (Ksp) for Ag2CrO4 is given by the following equation:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^(2-)(aq)
The expression for Ksp is:
Ksp = [Ag+]^2[CrO4^(2-)]
where [Ag+] and [CrO4^(2-)] are the concentrations of the silver ion and chromate ion in the equilibrium mixture, respectively.
To determine the value of Q, the reaction quotient, we need to determine the concentrations of Ag+ and CrO4^(2-) in the mixture of 5.00 mL of 0.0040 M AgNO3 and 4.00 mL of 0.0024 M K2CrO4. To do this, we need to make some assumptions:
1. The volumes of the two solutions are additive, so the total volume is 9.00 mL.
2. The AgNO3 and K2CrO4 solutions react completely to form Ag2CrO4.
First, we need to determine the moles of Ag+ and CrO4^(2-) in each solution:
For the AgNO3 solution:
moles of Ag+ = (0.0040 M) x (0.00500 L) = 2.0 x 10^-5 mol
For the K2CrO4 solution:
moles of CrO4^(2-) = (0.0024 M) x (0.00400 L) = 9.6 x 10^-6 mol
Since the AgNO3 and K2CrO4 react in a 1:1 ratio to form Ag2CrO4, the limiting reactant is K2CrO4. Therefore, all of the CrO4^(2-) is used up in the reaction, and the concentration of CrO4^(2-) in the equilibrium mixture is zero.
The concentration of Ag+ in the equilibrium mixture is:
[Ag+] = moles of Ag+ / total volume of mixture
[Ag+] = (2.0 x 10^-5 mol) / (9.00 x 10^-6 L)
[Ag+] = 2.22 M
Now, we can calculate the value of Q:
Q = [Ag+]^2[CrO4^(2-)] = (2.22 M)^2(0 M) = 0
Since Q is equal to zero and Ksp is greater than zero (1.8 x 10^-10), the reaction is not at equilibrium and Ag2CrO4 will precipitate from the solution.
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Select the correct answer.
where are globular clusters usually found?
oa. disk
ob. nucleus
oc. interstellar space
od. halo
Globular clusters are usually found in the halo of a galaxy. The answer is d.
Globular clusters are dense, spherical collections of stars that orbit a galactic center. They are typically composed of tens of thousands to hundreds of thousands of stars and are some of the oldest known objects in the universe.
Globular clusters are usually found in the halo of a galaxy, which is the outermost region of a galaxy that surrounds the disk.
This is because they are thought to have formed early in the history of the galaxy, when the halo was still being formed.
In contrast, stars in the disk of a galaxy are typically younger and more spread out, with less dense collections of stars. The nucleus of a galaxy is the central region, which usually contains a supermassive black hole and dense concentrations of stars.
Therefore, the correct answer is d. halo.
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which are true of the greenhouse effect? multiple select question. all energy from the sun is absorbed by atmospheric gases. some sunlight is absorbed and some is reflected by the atmosphere. some infrared energy is absorbed by gases such as carbon dioxide (co2), water vapor (h2o), and methane (ch4). it changes sunlight and transforms it into carbon dioxide. some light is absorbed by the land and oceans, which radiate infrared energy back into the atmosphere.
The true statements regarding the greenhouse effect are:
Some sunlight is absorbed and some is reflected by the atmosphere.Some infrared energy is absorbed by gases such as carbon dioxide (CO2), water vapor (H2O), and methane (CH4).Some light is absorbed by the land and oceans, which radiate infrared energy back into the atmosphere. Options B, C and E are correct.The greenhouse effect is a natural process that occurs when certain gases in the atmosphere, known as greenhouse gases, trap heat from the sun and prevent it from escaping back into space. This helps to keep the Earth's surface warm enough to support life. However, human activities, such as burning fossil fuels, have increased the concentration of greenhouse gases in the atmosphere, which has led to an enhanced greenhouse effect and global warming.
In the greenhouse effect, not all energy from the sun is absorbed by atmospheric gases. Rather, some of it is reflected back into space by the atmosphere. Additionally, the greenhouse effect does not change sunlight into carbon dioxide; rather, it is the burning of fossil fuels and other human activities that release carbon dioxide into the atmosphere. Options B, C and E are correct.
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1. From the chemicals listed on your lab handout, write down the weak acid (with its pKa) and its conjugate base that would create a buffer that best fits your protein. Would you expect for your buffer to have more acid or more base?
My assigned protein is Xylanase and has an optimum pH of 5. 5.
2. Buffers are used to the inhibit the change of pH upon the addition of strong acids and bases. If you were to add 0. 1 M HCl to your buffer, would you expect the pH to change? If so, would the pH increase or decrease? What would happen if 0. 1M NaOH were to be added instead?
3. Keeping your buffer composition from question 1 in mind, would you expect to use a larger volume of HCl or NaOH to change the pH of the buffer solution by one unit? Explain
1. Based on the information provided, a weak acid with a pKa close to the ideal pH of 5.5 would be the best buffer for Xylanase. Acetic acid (CH3COOH) can be utilised as the weak acid component of the buffer since it has a pKa of 4.76, which is close to the ideal pH.
Acetate ion (CH3COO-), its conjugate base, can also be utilised as a buffering agent. Since Xylanase prefers an acidic pH (below 7), we would anticipate the buffer to contain more acid than base.
2. The pH of the buffer would drop if 0.1 M HCl was introduced because the weak acid would arise when the H+ ions from the HCl react with the conjugate base in the buffer.
Instead, if 0.1 M NaOH was added, the pH would rise as the weak acid in the buffer reacts with the NaOH's OH- ions to form the conjugate base. The capacity of the buffer and the quantity of HCl or NaOH injected, however, would determine how much the pH changed.
3. It would take more HCl to raise the pH by one unit in the buffer in question 1 since it contains a weak acid and its conjugate base. This is due to the fact that the conjugate acid might be created when the weak acid component of the buffer reacts with extra H+ ions to limit significant pH shifts.
On the other hand, if NaOH were to be added to the buffer, the buffer's acid component would be consumed, causing a greater pH change.
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Design a concept map that shows the relationship between pressure, volume, and temperature in boyle's charles's and gay-lussac's laws
Gay-Lussac's Laws, also known as the Pressure-Temperature Law and Volume-Temperature Law, are a set of gas laws that explain the relationship between pressure, volume, and temperature in a given gas sample.
According to Gay-Lussac's Laws, if the pressure of a gas is increased while the volume remains constant, the temperature of the gas will also increase. Similarly, if the volume of a gas is decreased while the pressure remains constant, the temperature of the gas will increase as well.
In terms of a concept map, Gay-Lussac's Laws can be placed in the center with arrows pointing to both Boyle's Law and Charles's Law, which are two other gas laws that are also related to pressure, volume, and temperature.
Boyle's Law states that the pressure of a gas is inversely proportional to its volume, while Charles's Law states that the volume of a gas is directly proportional to its temperature.
To connect these three gas laws, the arrows can be labeled with key terms such as "pressure," "volume," and "temperature," with each gas law demonstrating how changes in one variable will affect the others.
The concept map can also include real-world examples of each gas law, such as how a tire pressure gauge can be used to demonstrate Boyle's Law, or how a hot air balloon can be used to demonstrate Charles's Law.
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If the solubility of CO2 is 0. 348 g/100 ml water at 101. 3 kPa, calculate the solubility of CO2 in water at a pressure of 263. 4 kPa. Assume the temperature is constant at 0°C
The solubility of CO₂ in water at 263.4 kPa is 1.064 g/100 ml water.
We can use Henry's law to calculate the solubility of CO₂ in water at a different pressure. Henry's law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Thus, we can set up the following equation:
(Solubility at new pressure) / (Solubility at reference pressure) = (Partial pressure of gas at new pressure) / (Partial pressure of gas at reference pressure)We are given the solubility of CO₂ at a reference pressure of 101.3 kPa, which is 0.348 g/100 ml water. We want to find the solubility at a new pressure of 263.4 kPa. We can rearrange the equation above to solve for the solubility at the new pressure:
(Solubility at new pressure) = (Partial pressure of gas at new pressure) x (Solubility at reference pressure) / (Partial pressure of gas at reference pressure)We know that the temperature is constant at 0°C, so we can assume that the solubility is directly proportional to the partial pressure. Thus, we can set up a ratio:
(Solubility at new pressure) / (Solubility at reference pressure) = (Partial pressure of gas at new pressure) / (Partial pressure of gas at reference pressure)Plugging in the given values, we get:
(Solubility at new pressure) / (0.348 g/100 ml) = (263.4 kPa) / (101.3 kPa)Solving for the solubility at the new pressure, we get:
Solubility at new pressure = (263.4 kPa) / (101.3 kPa) x (0.348 g/100 ml)Solubility at new pressure = 1.064 g/100 mlTherefore, the solubility of CO₂ in water at a pressure of 263.4 kPa is 1.064 g/100 ml water.
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Walking at a brisk pace burns off about 280 cal/h. how long would you have to walk to burn off the calories obtained from eating a cheeseburger that contained 25 g of protein, 25 g of fat, and 31 g of carbohydrates? [hint: one gram of protein or one gram of carbohydrate typically releases about 4 calg, while fat releases 9 cal/g. ] hours
You would need to walk at a brisk pace for about 1 hour and 40 minutes to burn off the calories obtained from eating the cheeseburger.
25 g of protein and 31 g of carbohydrates release 4 cal/g, which equals 240 calories. 25 g of fat release 9 cal/g, which equals 225 calories. So, the total calories in the cheeseburger are 465.
Now, to burn off 465 calories at a rate of 280 cal/h, we need to divide 465 by 280, which equals 1.66 hours or approximately 1 hour and 40 minutes.
In summary, to burn off the calories obtained from a cheeseburger containing 25 g of protein, 25 g of fat, and 31 g of carbohydrates, you would need to walk at a brisk pace for about 1 hour and 40 minutes.
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0. 18 g of a
divalent metal was completely dissolved in 250 cc of acid
solution containing 4. 9 g H2SO4 per liter. 50 cc of the
residual acid solution required 20 cc of N/10 alkali for
complete neutralization. Calculate the atomic weight of
metal.
39.
Ans: 36
0.18 g of a divalent metal was completely dissolved in 250 cc of acid solution containing 4. 9 g H₂SO₄ per liter. 50 cc of the residual acid solution required 20 cc of N/10 alkali for complete neutralization. The atomic weight of metal is 45 g/mol.
First, we need to determine the moles of H₂SO₄ present in 250 cc of the acid solution:
4.9 g/L = 0.0049 g/cc
0.0049 g/cc x 250 cc = 1.225 g of H₂SO₄
Next, we can calculate the number of moles of H₂SO₄ that were neutralized by the alkali solution:
20 cc of N/10 NaOH = 0.002 mol NaOH
Since the reaction is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
then 1 mol of H₂SO₄ reacts with 2 mol of NaOH, therefore 0.004 mol of H₂SO₄ reacted with 0.002 mol of NaOH.
So, the remaining number of moles of H₂SO₄ is:
0.004 mol - 0.002 mol = 0.002 mol
Now we can calculate the moles of metal present in the solution:
0.18 g / atomic weight = moles of metal
We can use the remaining H₂SO₄ to find the number of moles of metal:
1 mol of H₂SO₄ reacts with 1 mol of metal, so the number of moles of metal is equal to the number of moles of H₂SO₄ remaining:
0.002 mol H₂SO₄ = 0.002 mol metal
Now we can solve for the atomic weight:
0.18 g / 0.002 mol = 90 g/mol
Since the metal is divalent, we need to divide by 2 to get the atomic weight:
90 g/mol / 2 = 45 g/mol
Therefore, the atomic weight of the metal is 45 g/mol.
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The element krypton is a gas that shows almost no chemical activity. To find another element with similar properties, the student should look on the periodic table of the elements for an element _____.
In the same group
I’m the same period
With the same net charge
With the same atomic mass
To find another element with similar properties, the student should look on the periodic table of the elements for an element In the same group.
The student should look for an element in the same group as krypton on the periodic table of elements. Elements in the same group have similar chemical properties because they have the same number of valence electrons, which determine how the element interacts with other elements.
Krypton is a noble gas located in group 18 (also known as group 8A) of the periodic table. Other elements in this group, such as neon, helium, and xenon, also have very low chemical reactivity due to their full valence electron shells.
Elements in the same period as krypton may have similar atomic properties, but they are not guaranteed to have similar chemical properties since their valence electron configurations differ.
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If the pressure of a 7. 2 liter sample of gas changes from 735 torr to 800 torr and the temperature remains
constant, what is the new volume of the gas? (6. 62 L)
Answer:
you equate the question 800×7.2 divide the answer by 735.And you'll get 7.84litre then covert to 0.0m³ if the question says so to get 0.00784
What is the percent dissociation of HNO2 when 0. 058 of sodium nitrate is added to 110. 0ml of a 0. 060 M HNO solution? K, for HNO2 is 4. 0x10^-4
The percent dissociation of HNO₂ comes out to be 5.2% which is shown in the below secion.
The calculations of pKa is done as follows-
pKa = - log Ka
= - log (4.0 x 10⁻⁴)
= 3.398
Mole of NaNO₂ = mass / molar mass
= 0.058 g / 68.9953 g/mole
= 8.406 x 10⁻⁴ mole
Mole of HNO₂ = 0.110 L * 0.060 mole / L = 6.6 x10⁻³ mole.
Resulting solution is buffer solution.
pH = pKa + log [salt] / [acid]
Substituting the known values in the above formula.
pH = 3.398 + log ( 8.406 x 10⁻⁴ / 6.6 x 10⁻³ )
pH = 2.503
The pH can also be evaluated using the below expression.
pH = -log[H⁺]
-log[H] = 2.503
[H⁺]= 3.14 x 10⁻³ M
Thus
Percent of ionization = 3.14 x 10⁻³ M x 100 / 0.060 = 5.2 %
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How do you calculate the concentration of obtained solution with 2 solutions having the same concentration but different volume?
^
50.0cm³ of 0.0250 mol/dm³ nitric acid was mixed with 40.0 cm³ of 0.0250 mol dm/³ sulfuric acid.
Sketchpad
a chemist dilutes 2.0 l of a 1.5 m solution with water until the final volume is 6.0 l. what is
the new molarity of the solution?
show your work
The new molarity of the solution after dilution is 0.5 M.
To solve this problem, we can use the formula:
[tex]M_2 = M_1V_1 / V_2[/tex]
where [tex]M_1[/tex] and [tex]V_1[/tex] are the initial molarity and volume of the solution, and [tex]M_2[/tex] and [tex]V_2[/tex] are the final molarity and volume of the diluted solution.
In this case, we have:
[tex]M_1[/tex] = 1.5 M
[tex]V_1[/tex] = 2.0 L
[tex]V_2[/tex] = 6.0 L
We want to find the final molarity, [tex]M_2[/tex].
Using the formula, we can solve for [tex]M_2[/tex]:
[tex]M_2 = M_1V_1 / V_2[/tex]
Substituting the given values, we get:
[tex]M_2[/tex] = (1.5 M) × (2.0 L) / (6.0 L) = 0.5 M
Therefore, the new molarity of the solution is 0.5 M.
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Given the chart of bond energies, calculate the enthalpy change for the reaction below. Show all work
The enthalpy of the reaction can be obtained as 118 kJ/mol.
What is the reaction enthalpy?Reaction enthalpy, also known as heat of reaction or ΔHrxn, is the change in enthalpy that occurs during a chemical reaction. It is defined as the difference between the enthalpy of the products and the enthalpy of the reactants.
We have;
Enthalpy of reaction = Bonds broken - Bonds formed
Enthalpy of reaction = [4(413) + 2(495) - [2(799) + 2(463)
= [1652 + 990] - [1598 + 926]
=2642 - 2524
= 118 kJ/mol
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Explain why the following carboxylic acids cannot be prepared by a malonic ester synthesis. Part A A line-angle formula shows a ring with six vertices and alternating single and double bonds. A CH2COH group, with an O atom double-bonded to the second (from left to right) carbon atom, is attached to one of the ring vertices. A line-angle formula shows a ring with six vertices and alternating single and double bonds. A CH2COH group, with an O atom double-bonded to the second (from left to right) carbon atom, is attached to one of the ring vertices. An SN2 reaction cannot be done on benzyl bromide. An SN2 reaction cannot be done on bromobenzene. An SN2 reaction cannot be done on dibromobenzene. The bromide required for the synthesis is unstable
The first two carboxylic acids described contain a benzene ring, which is not susceptible to the malonic ester synthesis.
The malonic ester synthesis requires a compound with a methyl group adjacent to both carboxylate groups, and a benzene ring does not fulfill this requirement. The last two carboxylic acids described cannot be prepared by the malonic ester synthesis because an SN₂ reaction cannot be performed on compounds with bulky substituents or with two or more halogen atoms attached to the same carbon atom.
The synthesis requires the use of an alkyl halide that can undergo an SN₂ reaction with sodium ethoxide, but benzyl bromide, bromobenzene, and dibromobenzene are not suitable for this type of reaction. Additionally, the bromide required for the synthesis is unstable, which further complicates the reaction.
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A 20.0g sample of a hydrocarbon is found to contain 2.86g hydrogen. what is the percent by mass of carbon in the hydrocarbon
The percent by mass of carbon in the hydrocarbon is 85.7%.
To find the percent by mass of carbon in the hydrocarbon, follow these steps:
1. Determine the mass of hydrogen in the hydrocarbon: 2.86g.
2. Calculate the mass of carbon in the hydrocarbon by subtracting the mass of hydrogen from the total mass: 20.0g (total mass) - 2.86g (mass of hydrogen) = 17.14g (mass of carbon).
3. Calculate the percent by mass of carbon by dividing the mass of carbon by the total mass of the hydrocarbon, and then multiply by 100: (17.14g carbon / 20.0g total mass) * 100 = 85.7%.
So, the percent by mass of carbon in the 20.0g hydrocarbon sample is 85.7%.
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A sample of graphite with a mass of 15.0 grams drops from an initial temperature of 22°C to a
final temperature of 12°C. Calculate how much heat was transferred, and state whether it was
gained or lost based on the sign of your answer.
Answer:
106.5 J, and it was lost.
Explanation:
To calculate the amount of heat transferred, we can use the following formula:
Q = m * c * ΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
For graphite, the specific heat capacity is approximately 0.71 J/g°C.
So we have:
Q = 15.0 g * 0.71 J/g°C * (-10°C)
Q = -106.5 J
The negative sign of the answer indicates that the graphite lost heat, since its temperature decreased. Therefore, the heat was transferred from the graphite to its surroundings.
So the amount of heat transferred from the graphite was 106.5 J, and it was lost.
Question 1 (2 points)
2. 5 L of a gas is heated from 200 K to 300 K. What is the final volume of the gas?
The final volume of the gas is 3.75 L. This result can be explained by the fact that as the temperature of the gas increased, the kinetic energy of its particles also increased, causing them to move faster and occupy a larger volume.
According to the ideal gas law, PV = nRT, the volume of a gas is directly proportional to its temperature.
Therefore, if the temperature of a gas is increased while its pressure and amount remain constant, its volume will also increase.
In this case, the initial volume of the gas is 2.5 L and its temperature is increased from 200 K to 300 K. To find the final volume of the gas, we can use the following equation:
V2 = (T2/T1) x V1
where V1 is the initial volume of the gas, T1 is the initial temperature of the gas, T2 is the final temperature of the gas, and V2 is the final volume of the gas. Plugging in the values, we get:
V2 = (300 K/200 K) x 2.5 L
V2 = 3.75 L
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Which of the following solutions will have the greatest concentration?
a. 2 moles of solute dissolved in 1 liter of solution
b. 0.3 mole of solute dissolved in 0.6 liter of solution
c. 2 moles of solute dissolved in 10 liters of solution
d. 0.1 mole of solute dissolved in 0.5 liter of solution
2 moles of solute dissolved in 1 liter of solution has the greatest concentration.
What is concentration of a solution?Concentration refers to the quantity of solute that is dissolved in a specific amount of solution, and it is commonly measured in units such as moles per liter or grams per liter.
Equation:To determine which solution has the greatest concentration, we need to calculate the number of moles of solute present in each solution and then compare the values.
a. Concentration = 2 moles / 1 liter = 2 M
b. Concentration = 0.3 moles / 0.6 liters = 0.5 M
c. Concentration = 2 moles / 10 liters = 0.2 M
d. Concentration = 0.1 moles / 0.5 liters = 0.2 M
Comparing the concentrations, we see that solution (a) has the greatest concentration of 2 M, while the other solutions have concentrations of 0.5 M or lower.
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How many kj are released when 4.30 mol mg reacts with an excess of oxygen?
if 6.40 mol magnesium oxide are produced, how much energy is released?
if 68.9 g mg react with an excess of oxygen, how much energy is released?
the reaction produces 5,356 kj of energy. how many grams of mgo are formed?
The reaction of 4.30 mol of magnesium with an excess of oxygen produces 6.40 mol of magnesium oxide (MgO).
What is magnesium oxide ?Magnesium oxide is a white, odorless inorganic compound composed of magnesium and oxygen atoms. It is a strong basic oxide and an important mineral component of many rocks and soils. It has a wide range of industrial uses, such as in the production of cement, ceramics, and glass. It is also used as an antacid and laxative, and as a supplement to increase dietary magnesium intake.
The energy released in this reaction can be determined using the following equation:E = ΔHf (MgO) x (6.40 mol MgO)
In this equation, ΔHf (MgO) is the molar enthalpy of formation of magnesium oxide. The molar enthalpy of formation of magnesium oxide is -601.8 kJ/mol. Therefore, the total energy released in this reaction is:
E = -601.8 kJ/mol x (6.40 mol MgO)
E = -3,854.7 kJ.To determine the number of grams of MgO produced, we can use the following equation: Mass (MgO) = (6.40 mol MgO) x (Molar mass MgO) .
The molar mass of MgO is 40.3 g/mol. Therefore, the mass of MgO produced is: Mass (MgO) = (6.40 mol MgO.
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Is baking soda soluble in soda? Is sugar soluble in soda?
Baking soda is actually a compound known as sodium bicarbonate, which is water-soluble. Sugar, on the other hand, is also soluble in water and other liquids that contain water.
This means that it dissolves in water and can also dissolve in other liquids that contain water, such as soda. Therefore, baking soda is indeed soluble in soda.
Sugar, on the other hand, is also soluble in water and other liquids that contain water. This includes soda, which is a carbonated beverage that typically contains a high amount of dissolved sugar.
However, the solubility of sugar in soda can depend on various factors such as the temperature of the soda, the amount of sugar present, and the type of sugar used.
In general, both baking soda and sugar are soluble in soda and can dissolve to some extent. However, the exact degree of solubility can vary depending on various factors. It is worth noting that excessive consumption of sugary soda can have negative impacts on health, so it is important to consume such beverages in moderation.
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Help what’s the answer?
The mass of the zinc hydroxide that we need for the reaction is about 21.8 g.
What is the equation of reaction?The equation of a reaction is a chemical equation that represents the chemical change that occurs during a chemical reaction. It is typically written in the form:
Reactants → Products
where the reactants are the starting materials and the products are the substances that are formed as a result of the reaction.
The equation of the reaction is;
Zn(OH)2 + H2SO4 → ZnSO4 + 2H2O
Number of moles of H2SO4 = 21.1 g/98 g/mol
= 0.22 moles
If the reaction is 1:1,
Mass of the Zn(OH)2 required = 0.22 moles * 99 g/mol
= 21.8 g
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How many grams of CaCl2 would be required to produce a. 750 M solution with a 855 ml volume?
8.32 grams of CaCl2 are required to produce 750 ml of a 0.100 M CaCl2 solution.
The number of moles of solute that can dissolve in 1 L of a solution is known as molarity or molar concentration.
Volume of solution (in litres) / Number of solutes (in moles), or
C = n / V
According to the question,
V = 750 ml and C = 0.100 M
Let's convert millilitres to litres for this.
We know 1 L = 1000 ml
Consequently, 750 ml equals (750/1000) L, or 0.75 L.
So, V = 0.75 L
We know that C = n / V
So, n = C x V
n = 0.100 x 0.75 = 0.075
The solute contains n moles in total.
CaCl2 thus has a mole count of 0.075 moles.
In 750 ml of solution, this demonstrates that there are 0.075 moles of CaCl2.
We must know the molar mass of CaCl2 in order to calculate the mass of CaCl2 in grams.
To do this, we must use the periodic table to determine the atomic masses of each atom.
CaCl2 consists of 1 Ca and 2 Cl atoms.
Atomic mass of Ca is 40.08 g and that of Cl is 35.45, so 2 x 35.45 = 70.90 g.
We obtain the mass of CaCl2 in grams by averaging these measurements.
Hence, mass of CaCl2 = 40.08 + 70.90 = 110.98 g
Thus, 110.98 g equals 1 mole of CaCl2.
Therefore, 0.075 moles of CaCl2 will weigh 8.3235 g, which is rounded to 8.32 g, or 0.075 x 110.98 g.
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The complete question is
How many grams of calcium chloride will be needed to make 750 mL of a 0.100 M CaCl2 solution?
Substances can be the same if one has more of the same type of repeating group of atoms
Yes, substances can be the same if one has more of the same type of repeating group of atoms.
For example, polymers are made up of repeating units of the same monomer, and the number of monomers can vary, resulting in different sizes of polymers but still the same substance. Another example is isotopes, which are elements with the same number of protons but varying numbers of neutrons.
They have the same chemical properties and can form the same compounds despite having different atomic masses. Thus, substances can be identical in terms of their chemical properties even if they have different physical properties due to variations in the number of repeating groups of atoms or isotopes.
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What is the name of this branched alkene? Please help me as fast as possible I need to study, please! ILL MARK AS BRAINIEST FOR CORRECT ANSWER PLEASE HELP ITS DUE IN 25 MINUTES
6-ethyl-8-methyl-5-propyl- non-2-ene is the name of this branched alkene. The World Union of Applied and Pure Chemistry (IUPAC).
The World Union of Applied and Pure Chemistry (commonly abbreviated to IUPAC) recommends a systematic approach for the terminology of organic compounds, which is referred to as the IUPAC classification of organic compounds.
The IUPAC naming criteria are occasionally followed by chemists, nevertheless, as some compounds have names that are incredibly long and difficult to pronounce. More banal names are given to these substances. 6-ethyl-8-methyl-5-propyl- non-2-ene is the name of this branched alkene.
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Calculate the molar solubility of agbr in 2.8×10−2 m agno3 solution. the ksp of agbr is 5.0 * 10-13
The molar solubility of [tex]AgBr[/tex] in [tex]2.8 x 10^-2 M AgNO3[/tex] solution is [tex]7.1 x 10^-7 M[/tex].
To calculate the molar solubility of [tex]AgBr[/tex] in [tex]2.8 x 10^-2 M AgNO3[/tex] solution, we need to use the common ion effect. The [tex]Ag+[/tex] ion is a common ion in both [tex]AgBr and AgNO3[/tex]. When we add [tex]AgNO3[/tex] to a solution containing AgBr, it adds more [tex]Ag+[/tex] ions to the solution and causes a shift in the equilibrium to the left. The solubility of [tex]AgBr[/tex]decreases due to this effect.
The balanced equation for the dissolution of [tex]AgBr[/tex] is:
[tex]AgBr(s) ⇌ Ag+(aq) + Br-(aq)[/tex]
The Ksp expression for AgBr is:
Ksp = [Ag+][Br-] = 5.0 x 10^-13
Let x be the molar solubility of [tex]AgBr[/tex]in [tex]2.8 x 10^-2 M AgNO3[/tex]solution. Then the concentration of [tex]Ag+[/tex] ion is[tex][Ag+] = 2.8 x 10^-2 + x[/tex], and the concentration of [tex]Br-[/tex] ion is[tex][Br-] = x[/tex].
Substituting these values into the Ksp expression, we get:
[tex]Ksp = (2.8 x 10^-2 + x)(x) = 5.0 x 10^-13[/tex]
Simplifying the equation and neglecting x in comparison to [tex]2.8 x 10^-2[/tex], we get:
[tex]x^2 = 5.0 x 10^-13x = sqrt(5.0 x 10^-13) = 7.1 x 10^-7 M[/tex]
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According to the law of conservation of mass, the total mass of the products of a chemical reaction should equal the total mass of the reactants. In our reaction on a balance, however, the mass went down, from 253. 0 g to 250. 2 grams. Using the given chemical equation, explain what caused the apparent loss of mass
The apparent loss of mass from 253.0 g to 250.2 g can be attributed to the involvement of a gas in the chemical reaction, either as a product or a reactant. This gas escapes the system during the reaction, causing a decrease in the observed mass, but the law of conservation of mass still holds true as the total mass is conserved in the reaction.
According to the law of conservation of mass, the total mass of the products of a chemical reaction should equal the total mass of the reactants. In your reaction, the mass went down from 253.0 g to 250.2 g, which seems to contradict this law. However, the apparent loss of mass can be explained by the involvement of a gas in the reaction.
Here's a step-by-step explanation:
1. Identify the given chemical equation. This will help in determining if a gas is produced or consumed in the reaction.
2. Examine the reactants and products to see if any of them are gases. Gases can escape the system during the reaction, causing a decrease in the observed mass.
3. If a gas is produced, this explains the apparent loss of mass. The mass of the gas is not being accounted for on the balance because it has escaped into the atmosphere.
4. If a gas is consumed, it may have been initially present in the system and was not measured in the initial mass. Once it is consumed, the mass of the system would appear to decrease.
In summary, the apparent loss of mass from 253.0 g to 250.2 g can be attributed to the involvement of a gas in the chemical reaction, either as a product or a reactant.
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