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3 years ago

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A 2.5-liter sample of a gas has 0.30 mole of the gas. If 0.15 mole of the gas is added, what is the final volume of the gas? Tem

perature and pressure remain constant. 3.4 liters 3.8 liters 4.2 liters 4.7 liters

1 answer:

alisha [4.7K]3 years ago

8 0

Answer:

It is 4.2 liters

Explanation:

Have an amazing day!

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Which focusses on the composition of matter?

Gwar [14]

Answer:

Inorganic chemistry: Inorganic chemistry is the study of chemicals that, in general, are not primarily based on carbon. Analytical chemistry: Analytical chemistry is the study of the composition of matter. It focuses on separating, identifying, and quantifying chemicals in samples of matter.

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3 years ago

What element is chemically combined with hydrogen

Fudgin [204]

Uh I think water is??

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3 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

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4 years ago

What is the chemical that had polluted flints drinking water supply

Ber [7]

Answer:

Lead

Explanation:

While taking sample from flint's distillery, theybfoind 13000 ppb or parts per billion lead in the water itself

5 0

3 years ago

To what temperature in kelvin will a 2.3 L balloon have to be heated to expand to a volume of 400 L Assume the initial temperatu

olchik [2.2K]

Answer:

T₂ = 51826.1 K

Explanation:

Given data:

Initial Volume = 2.3 L

Final volume = 400 L

Initial temperature = 25 °C (25+ 273 = 298 K)

Final temperature = ?

Solution:

V₁/T₁ = V₂/T₂

T₂ = V₂ T₁/V₁

T₂ = 400 L . 298 K / 2.3 L

T₂ = 119200 K. L / 2.3 L

T₂ = 51826.1 K

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3 years ago