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2 years ago

During an experiment, 575 mL of neon gas at 101 kPa were compressed in a cylinder to a volume of 144 mL.

Chemistry

1 answer:

Ne4ueva [31]2 years ago

8 0

Answer:

403.3 kPa is the new pressure

Explanation:

This problem is solved by this formula:

P₁ . V₁ = P₂ . V₂

101kPa . 575 mL = P₂ . 144mL

(101kPa . 575 mL) / 144 mL = P₂

403.3 kPa = P₂

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2 years ago

Can somebody help me, please!

zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

Rubidium-85

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

72/100= 0.72 or 72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

Rubidium-87

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

28/100= 0.28 or 28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

Atomic Mass of Rubidium:

Add the two numbers together.

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2 years ago

Anyone can help me in science assignment

olya-2409 [2.1K]

Answer:

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Explanation:

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Read 2 more answers

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

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2 years ago