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krek1111 [17]
3 years ago
12

A sample of helium gas has a volume of 0.180L, a pressure of 0.800a and a temperature of 29 C. What is the new temp or gas of vo

lume of 90mL and pressure of 3.20 atm
Chemistry
1 answer:
Naddik [55]3 years ago
7 0

Answer:

The new temperature of the gas is 604K

Explanation:

Assuming standard temperature and pressure, to calculate the temperature of the gas we use the general gas equation;

Step 1 :write the general gas equation

P1V1/ T1 = P2V2/T2

Step 2: Write out the values, covert the necessary values to the standard values.

P1 = 0.800atm.

V1 = 0.180L

T1 = 29°C = 273 + 29 = 302K

P2 , 3.20atm

V2 = 90mL = 90 * 10^-3L = 0.09L

Step 3: Solve for T2

The new temperature T2 of the gas is:

T2 = P2V2T1/ P1V1

T2 = 3.20 * 0.09 * 302 / 0.800 * 0.180

T2 = 86.976 / 0.144

T2 = 604K

The new temperature is of the gas is 604K.

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Answer:

The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .

6 0
3 years ago
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

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4 years ago
What is true about resonance structures? Select the correct answer below: They are all different molecules. The interconvert wit
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Answer:

The correct answer is: <em>They each partially describe the bonding in a molecule.</em>

Explanation:

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Answer:

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