Ask question

Ask question

All categories

3 years ago

7

A sample of xenon gas at a pressure of 1.14 atm and a temperature of 20.4 °C, occupies a volume of 16.9 liters. If the gas is al

lowed to expand at constant temperature to a volume of 27.7 liters, the pressure of the gas sample will be atm.

1 answer:

kondor19780726 [428]3 years ago

8 0

Answer:

P2= 0.696atm

Explanation:

Applying Boyle's law

P1V1= P2V2

1.14×16.9= P2× 27.7

Simplify

P2= (1.14×16.9)/27.7

P2= 0.696atm

You might be interested in

What is another name for a homogeneous mixture?

statuscvo [17]

The answer is a solution im pretty sure <3

4 0

3 years ago

What is true of neutrons? They have no charge and are located inside the nucleus. They have no charge and are located outside th

Inessa [10]

They have no charge and are located inside the nucleus

3 0

3 years ago

Read 2 more answers

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

3 years ago

Apresentando a configuração eletrônica 1s2 2s2 2p6 3s2 3p4 um determinado elemento químico se mostra com forte tendência para:

Bess [88]

Answer:

E

Explanation:

To know what will happen, we need to know

the total number of electrons the element have.

This can be calculated by adding all the electrons together

That would be 2 + 2 + 6 + 2 + 4 = 16

This shows that the element is sulphur.

Now using the KLMN electronic configuration, sulphur has the shells 2,8,6 meaning there are 2 extra electrons needed for octet configuration to be achieved.

This makes option E the correct answer to the question

6 0

2 years ago

Pesticides Physical properties or Chemical Properties

larisa86 [58]

A chemical Property I hope this helps friend ^^

8 0

3 years ago