How much energy is needed to raise the temperature of 125g of water from 25.0oC to 35.0oC? The specific heat of water is 4.184
1 answer:
You might be interested in
<u>Answer:</u>
<u>For 1:</u> The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg
<u>For 2:</u> The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608
<u>Explanation:</u>
<u>For 1:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For helium:</u>
Given mass of helium = 0.723 g
Molar mass of helium = 4 g/mol
Putting values in equation 1, we get:
- <u>For methane gas:</u>
Given mass of methane gas = 3.43 g
Molar mass of methane gas = 16 g/mol
Putting values in equation 1, we get:
To calculate the mole fraction , we use the equation:
.......(2)
To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:
......(3)
- <u>For Helium gas:</u>
We are given:
Putting values in equation 2, we get:
Calculating the partial pressure by using equation 3, we get:
Putting values in equation 3, we get:
- <u>For Methane gas:</u>
We are given:
Putting values in equation 2, we get:
Calculating the partial pressure by using equation 3, we get:
Putting values in equation 3, we get:
Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg
- <u>For 2:</u>
We are given:
Partial pressure of nitrogen gas = 363 mmHg
Partial pressure of carbon dioxide gas = 564 mmHg
Total pressure = (363 + 564) mmHg = 927 mmHg
Calculating the mole fraction of the gases by using equation 3:
<u>For nitrogen gas:</u>
<u>For carbon dioxide gas:</u>
Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608
Answer : The mass of oxygen will be, 36.8 grams
Solution : Given,
Moles of water = 2.30 moles
Molar mass of = 32 g/mole
First we have to calculate the moles of oxygen.
The balanced chemical reaction will be,
From the balanced reaction we conclude that
2 moles of water decomposes to give 1 mole of
2.30 moles of water decomposes to give moles of
Now we have to calculate the mass of oxygen.
Therefore, the mass of oxygen will be, 36.8 grams
Answer: HCl is the limiting reactant and the theoretical yield is 2.72 g of CO2. If the actual yield was 2.50 g then, the percent yield is 92.0% when rounding off is done only for the final answer.
Further Explanation:
In order to determine the theoretical yield and the percent yield of CO2, the following steps must be done:
- Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form.
- Determine the theoretical yield for CO2 when the limiting reactant is used.
- Get the percent yield by getting the ratio of the actual yield stated in the problem and the calculated theoretical yield multiplied by 100.
Determining the Limiting Reactant
The Limiting Reactant (LR) will produce fewer moles of the products. To check which of the reactants HCl or CaCO3 is the LR, we do dimensional analysis:
For HCl:
For CaCO3:
Since HCl produces fewer moles of CO2, then it is the limiting reactant. We will use the given amount to determine the theoretical yield for CO2.
Determining the Theoretical Yield
From Step 1, we know that 0.0617098 moles of CO2 will be produced. We will just convert this to grams.
Since the answer only requires 3 significant figures, the final answer is 2.72 grams CO2.
Determining the Percent Yield
Dividing the actual yield by the theoretical yield will give us the percent yield, which is an indicator of how efficient the experiment or the method used was.
From the problem, the actual yield was 2.50 g, hence, the percent yield is:
Rounding off to three significant figures, the percent yield is 92.0%. This suggests that the method used is somewhat efficient in producing CO2.
Learn More
- Learn More about Limiting Reactant brainly.com/question/7144022
- Learn More about Excess Reactant brainly.com/question/6091457
- Learn More about Stoichiometry brainly.com/question/9743981
Keywords: stoichiometry, theoretical yield, actual yield
Answer:
a. Adding a solute to a solvent results in freezing point depression that is lowering of the freezing point of the solvent
The equation that describes the effect is
b. The effect of adding a solute to a solvent have on the boiling point of the solvent is the elevation of the boiling point of the solvent
The equation that describes the effect of boiling point elevation is given as follows;
Explanation:
Where:
=
-
= Constant for Molal freezing point
i = van't Hoff factor which is the number of irons present in each solute molecule
m = Molality of the solution
Where:
=
-
= Constant for Molal freezing point
i = van't Hoff factor which is the number of irons present in each solute molecule
m = Molality of the solution