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2 years ago

2 moles of a gas have a pressure of 2 atm at a temperature of 300 K. What is the volume of the gas?

Chemistry

1 answer:

qwelly [4]2 years ago

4 0

No of moles=2mol=n
Pressure=2atm=P
Temperature=300K=T
Volume=V

$\\ \tt\hookrightarrow PV=nRT$

$\\ \tt\hookrightarrow V=\dfrac{nRT}{P}$

$\\ \tt\hookrightarrow V=\dfrac{2(8.314)(300)}{2}$

$\\ \tt\hookrightarrow V=8.314(300)$

$\\ \tt\hookrightarrow V=2494.2mL$

$\\ \tt\hookrightarrow V=24.94dm^3$

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I don’t understand this question and need help!!

Dafna1 [17]

Medium about 3 second? Not sure lol just need more points honestly lol

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2 years ago

10) How many grams are there in 1.00 x 10^24 molecules of BC13?

lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0

3 years ago

A 15.0 g piece of graphite is heated to 100.0°C and placed in a calorimeter. The

LenKa [72]

Answer:

0.714Jg^-10C^-1.

Explanation:

4 0

3 years ago

Read 2 more answers

If the volume occupied by the air in a bicycle pump is 525 cm3, and the pressure changes from 73.2 kPa to 122.5 k.Pa as the pist

Musya8 [376]

Answer:

$V_2=313.71\ cm^3$

Explanation:

Given that,

Initial volume, $V_1=525\ cm^3$

The pressure changes from 73.2 kPa to 122.5 k.Pa.

We need to find the new volume occupied by the air. Let it is V₂. It can be calculated using Boyle's law such that,

$P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{73.2\times 525}{122.5}\\\\V_2=313.71\ cm^3$

So, the new volume is $313.71\ cm^3$ .

6 0

2 years ago

A solution of acetic acid, CH3CO₂H(aq), is at equilibrium. How would the

s344n2d4d5 [400]

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation. (Option C)

<h3>How do systems at equilibrium respond to perturbation?</h3>

When a system at equilibrium suffers a perturbation, it shifts its equilibrium position to counteract such perturbation.

Let's consider a solution of acetic acid at equilibrium.

CH₃CO₂H(aq) = CH₃CO₂⁻(aq) + H⁺(aq)

If more acetic acid were added to the solution, the system will shift toward the products to counteract such an increase.

How would the system change if more acetic acid were added to the solution?

A. [H⁺] would decrease and [CH₃CO₂⁻] would increase. NO.

B. [H⁺] and [CH₃CO₂⁻] would decrease. NO.

C. [H⁺] and [CH₃CO₂⁻] would increase. YES. Both products would increase.

D. [H⁺] would increase and [CH₃CO₂⁻] would decrease. NO.

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation.

Learn more about equilibrium here: brainly.com/question/2943338

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2 years ago