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PtichkaEL [24]
3 years ago
13

A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the vo

lume be at a pressure of 463.0 mm Hg and a temperature of 777.0 K? The combined gas law equation is given below. P1V1 T1 = P2V2 T2 In this equation, P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
Chemistry
1 answer:
kozerog [31]3 years ago
6 0

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

P1 = 717 mmHg

T1 = 315 K

V2 = ?

P2 = 463 mmHg

T2 = 777 K

The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.

(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

0.346*777 = (463*V2) / (777)*777

268.842 = 463*V2

268.842/463 = (463*V2)/463

V2 = 0.581

Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

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When cyclohexene is treated with KMnO4, H2O, the syn-1,2-diol is produced. What reaction occurs when benzene is similarly treate
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Answer:

No reaction is observed

Explanation:

The benzene ring is aromatic. Being an aromatic ring, the benzene ring is remarkably stable to all reactions that destroy the aromatic ring.

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The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b
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Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

\Delta S_{vap}=\frac{\Delta H_{vap}}{T}

We can solve for the temperature as follows:

T=\frac{\Delta H_{vap}}{\Delta S_{vap}}

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3 years ago
A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature a
adell [148]

Answer:

volume of gas = 9.1436cm³

Explanation:

We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.

Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.

V1 =10.1cm³ , P1 =746mmHg,     T1=23°C =23+273=296k

V2 =? ,   P2 =760mmmHg ,     T2=0°C = 0+273 =273K

Using the combined gas equation to calculate for V2;

\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}

V2 =\frac{10.1*746*273}{760*296}

V2=9.1436cm³

6 0
3 years ago
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