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Maru [420]
3 years ago
6

How many moles of gas at 58° C does it take to fill a 1.75 L flask to a pressure of 12.5 kPa? Please help I will give brainliest

Chemistry
1 answer:
marin [14]3 years ago
3 0

Answer:

0.008 moles of gas are present

Explanation:

Given data:

Volume of gas = 1.75 L

Number of moles =  ?

Temperature of gas = 58°C

Pressure of gas = 12.5 KPa

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

58+273 = 331 K

Pressure = 12.5/101 = 0.12 atm

by putting values in formula:

0.12 atm× 1.75 L = n× 0.0821 atm.L/ mol.K   ×331 K

0.21 atm.L = n× 27.17atm.L/ mol

n = 0.21 atm.L /27.17atm.L/ mol

n = 0.008 mol

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5. A gas occupies 2000. Lat 100.0 K and exerts a pressure of 100.0 kPa. What volume will
Anna [14]

Answer:

4000 L

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial volume (V1) = 2000 L.

Initial temperature (T1) = 100 K.

Initial pressure (P1) = 100 kPa.

Final temperature (T2) = 400 K.

Final pressure (P2) = 200 kPa.

Final volume (V2) =..?

Step 2:

Determination of the new volume of the gas.

The new volume of the gas can be obtained by using the general gas equation as follow:

P1V1/T1 = P2V2/T2

100 x 2000/100 = 200 x V2/400

Cross multiply to express in linear form.

100 x 200 x V2 = 100 x 2000 x 400

Divide both side by 100 x 200

V2 = (100 x 2000 x 400)/(100 x 200)

V2 = 4000 L

Therefore, the new volume of the gas is 4000 L

5 0
3 years ago
The chemical formula for lithium fluoride is lif . a chemist determined by measurements that 0.035 moles of lithium fluoride par
Mnenie [13.5K]
Answer:
mass = 0.907865 grams

Explanation:
From the periodic table:
molar mass of Li = 6.941 grams
molar mass of F = 18.998 grams
Therefore:
molar mass of LiF = 6.941 + 18.998 = 25.939 grams/mole

number of moles can be calculated as follows:
number of moles = mass / molar mass
We have:
number of moles = 0.035 moles
molar mass = 25.939 grams/mole

Substitute in the equation to get the mass as follows:
0.035 = mass / 25.939
mass = 0.035 * 25.939 = 0.907865 grams

Hope this helps :)
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3 years ago
Which example provide the most complete descripition of an object motion? A. the bird watcher follwed the south trailer of five
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The answer is C.) hoped this helps
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Read 2 more answers
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S
Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of ^{235}UF_6=349.034348g/mol

Molar mass of ^{238}UF_6=352.041206g/mol

By taking their ratio, we get:

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}

From the above relation, it is clear that rate of effusion of ^{235}UF_6 is faster than ^{238}UF_6

Difference in the rate of both the gases, Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816

To calculate the percentage increase in the rate, we use the equation:

\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100

Putting values in above equation, we get:

\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%

The above calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

3 0
3 years ago
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