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Fed [463]
3 years ago
6

How many atoms are in the 1.5 moles of tin atoms?

Chemistry
1 answer:
Anastasy [175]3 years ago
6 0

Answer:

option C = 9.0 x 10²³ atoms

Explanation:

Data Given:

no. of moles of tin (Sn) atoms = 1.5 moles

no. of tin (Sn) atoms = ?

Solution:

Formula used to find number of atoms

                  no. of moles = no. of atoms / Avogadro's number

Rearrange the above equation:

          no. of atoms =   no. of moles x Avogadro's number . . . . . . (1)

Where

Avogadro's number = 6.022 x 10²³

Put values in equation 1

           no. of atoms = 1.5 x 6.022 x 10²³

           no. of atoms = 9.033 x 10²³

Round the figure = 9.0 x 10²³ atoms

So option C is correct

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Show the calculation of the mass (grams) of a 14.5 liter gas sample with a molecular weight of 82 when collected at 29°C and 740
mote1985 [20]

Answer:

In this conditions, the gaswll weight 46.74 g.

Explanation:

The idal gas law states that:

PV = nRT,

P: pressure = 740 mmHg = 0.97 atm

V: volume = 14.5 L

n: number of moles

R: gas constant =0.08205 L.atm/mol.K

T: temperature = 29°C = 302.15K

n = \frac{PV}{RT} \\n = \frac{0.97x14.5}{0.082 x 302.15 } \\n = 0.57 mol

1 mol gas ___ 82 g

0.57 mol gas __ x

x = 46.74 g

4 0
2 years ago
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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