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3 years ago

If the pressure of a 2.3 mole sample of gas is initially 1.5 atm, the volume is

Chemistry

1 answer:

andreyandreev [35.5K]3 years ago

7 0

Answer: last one
300 L
I hope this helps you

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A 8.65-L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.205 atm

Vsevolod [243]

The total pressure = 1.402 atm

<u><em>calculation</em></u>

Total pressure = partial pressure of gas A + partial pressure of gas B + partial pressure of third gas

partial pressure of gas A= 0.205 atm

Partial pressure of gas B =0.658 atm

partial pressure for third gas is calculated using ideal gas equation

that is PV=nRT where,

p(pressure)=? atm

V(volume) = 8.65 L

n(moles)= 0.200 moles

R(gas constant)=0.0821 L.atm/mol.k

T(temperature) = 11°c into kelvin =11+273 =284 k

make p the subject of the formula by diving both side by V

p =nRT/v

p = [(0.200 moles x 0.0821 L.atm/mol.K x 284 K)/8.65L)] =0.539 atm

Total pressure is therefore = 0.205 atm +0.658 atm +0.539 atm

=1.402 atm

6 0

3 years ago

What is the mass of two moles of the diatomic gas, N2

elena-s [515]

Answer:

$m_{N_2}=56.04 gN_2$

Explanation:

Hello!

In this case, since nitrogen diatomic gas is a molecule that has the following molar mass:

$M_{N_2}=14.01*2\\\\M_{N_2}=28.02g/mol$

Thus, since we have two moles of this molecule, we can compute the referred mass as shown below:

$m_{N_2}=2molN_2*\frac{28.02gN_2}{1molN_2}\\\\m_{N_2}=56.04 gN_2$

Best regards!

4 0

3 years ago

Select ALL factors in conservation: *

il63 [147K]

Answer:

ALL of these are factors in conservation.

Explanation:

8 0

3 years ago

Need help !!!!! ASAP

Nat2105 [25]

<h2>Hello!</h2>

The answer is:

The temperature will be the same, 37°C.

Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.

The law establishes the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

Then, we are given the following information:

$V_{1}=200mL\\P_{1}=4atm\\T_{1}=37\°C\\V_{2}=400mL\\P_{2}=2atm$

So, isolating the new temperature and substituting the given information, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\T_{2}=P_{2}V{2}*\frac{T_{1}}{P_{1}V_{1}} \\\\T_{2}=2atm*400mL*\frac{37\°C}{4atm*200mL}=37\°C$

Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.

The temperature will be the same, 37°C

Have a nice day!

4 0

3 years ago

What are the two basic properties of matter

NISA [10]

Physical and chemical properties.

7 0

4 years ago