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aivan3 [116]
3 years ago
14

A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the

syringe be heated/cooled in order to have a volume of 435 mL at 2.50 atm? Group of answer choices
Chemistry
1 answer:
Ivahew [28]3 years ago
8 0

<u>Answer:</u> The temperature to which the gas in the syringe must be heated is 720.5 K

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=1.88atm\\V_1=285mL\\T_1=355K\\P_2=2.50atm\\V_2=435mL\\T_2=?K

Putting values in above equation, we get:

\frac{1.88atm\times 285mL}{355K}=\frac{2.50atm\times 435mL}{T_2}\\\\T_2=\frac{2.50\times 435\times 355}{1.88\times 285}=720.5K

Hence, the temperature to which the gas in the syringe must be heated is 720.5 K

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In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters
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<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol

For the given chemical reaction:

Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = \frac{1}[1}\times 0.0851=0.0851mol of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Putting values in above equation, we get:

733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L

Hence, the volume of hydrogen gas collected over water is 2.13 L

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Answer:

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