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1 year ago

5. What is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm?

Chemistry

1 answer:

Helga [31]1 year ago

3 0

70.33 L is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm.

<h3>What is volume?</h3>

Volume is the percentage of a liquid, solid, or gas's three-dimensional space that it occupies.

Liters, cubic metres, gallons, millilitres, teaspoons, and ounces are some of the more popular units used to express volume, though there are many others.

We will use ideal gas law to find the volume

PV = nRT

Can also be written as

V = (nRT)/P

Where,

P = pressure

V = volume

n = amount of substance

R = ideal gas constant

T = temperature

Here, we have given

P = 3.5 atm

V = to find

n = 10 moles

R = 0.08206 L⋅atm/K⋅mol

T = 300k

Lets substitute the values

V = (10 × 0.08206 × 300)/3.5

V = 70.33 L

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A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0

3 years ago

Which of the following is NOT a reason why chemical bonds are important? *

KengaRu [80]

Answer:

the answer would be B :)

3 0

4 years ago

Aqueous acetic acid is neutralized by aqueous barium hydroxide. express your answer as a balanced molecular equation. identify a

Studentka2010 [4]

The balanced molecular equation for the given reaction is:

$2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).$

<h3>FURTHER EXPLANATION</h3>

To check that the equation is balanced count how many atoms are present for each element in the reactant and product side. If they are the same before and after reaction for all elements, then the reaction is deemed balanced.

The atom counting for the equation is shown below:

$2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).$

<u>Reactants </u> → <u>Products</u>

C: (2 x 2) =4 C: (2 x 2) = 4

H: (2 x 4) + (1 x 2) =10 H: (3 x 2) + (2 X 2) = 10

O: (2 x 2) + (1 x 2) = 6 O: (2 x 2) + (2 x 1) = 6

Ba: 1 Ba: 1

Since the number of atoms of each element are similar in the reactants and products, the equation is balanced.

To determine the state of the substances, consider their solubility.

The reactants are both aqueous (aq) as indicated in the problem.

The first product, $Ba(CH_3COO)_2$ is aqueous (aq) because based on the solubility rule, it will dissolve in water. Acetates are generally soluble.