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3 years ago

Which of the following could be absent from the grief process

Chemistry

1 answer:

grin007 [14]3 years ago

4 0

1. C.
2. A.
3. A. or B.
4. D.
5. B.
6. B.

Sorry if they are wrong. I tried my best.

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Need help !

yulyashka [42]

B is the answer loll

6 0

2 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

A sample of a gas has a volume of 852 mL at 298 K. If the gas is cooled to 200K, what would the new volume be?

Colt1911 [192]

Answer:

571.81 mL

Explanation:

Assuming constant pressure, we can solve this problem by using Charles' law, which states that at constant pressure:

V₁T₂=V₂T₁

Where in this case:

V₁ = 852 mL

T₂ = 200 K

V₂ = ?

T₁ = 298 K

We input the data:

852 mL * 200 K = V₂ * 298 K

And solve for V₂:

V₂ = 571.81 mL

The new volume would be 571.81 mL.

4 0

3 years ago

QUESTIONS

vagabundo [1.1K]

Answer:

Explanation:

3 0

3 years ago

How electrons can be arranged in atoms orbitals

Nuetrik [128]

electrons are arranged in shells around an atom's nucleus. Electrons closest to the nucleus will have the lowest energy.

3 0

4 years ago