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jeyben [28]
3 years ago
6

100 POINTS!!!!

Chemistry
2 answers:
11111nata11111 [884]3 years ago
8 0

fith statement

sixth statement

first statement I think

IrinaVladis [17]3 years ago
7 0
First statement
Fifth statement
Sixth statement
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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

3 0
3 years ago
1.What is another term for the 24 hour clock?
s344n2d4d5 [400]

Answer:

military time

Explanation:

military time

5 0
2 years ago
What is the result of having two batteries on an electric motor
aleksley [76]

Answer:

Well bro lemme tell u,

by connecting batteries, you can increase the voltage, amperage, or both.

Explanation:

8 0
2 years ago
Which of the following is not a physical property of a liquid?
Usimov [2.4K]
Gas is not a liquid, hope this helps, o and next time you might wanna put the answer choices thanks
7 0
3 years ago
An inexpensive and accurate method of measuring the quantity of electricity flowing through a circuit is to pass the current thr
Paha777 [63]

Total of 127.013 C of charge is passed

Given

 weight of Ag solution before current has passed = 1.7854 g

  weight of Ag solution after current has passed = 1.8016 g

  Molecular mass of Ag = 107.86 g

  Faraday's Constant = 96485

First of all we have to apply Faraday's First Law of Electrolysis i.e

         m = ZQ

  where

Z is propotionality constant (g/C)

Q is charge (C)

Hence,

 Z = Atomic mass of substance/ Faraday's Constant

    = \frac{107.86 g}{96485 C}

    = 0.0011178 g/C

Now ,

   change in mass before and after the passing of current (Δm)

       Δm = 1.8016g-1.7854g

             =   0.0162g

    Now  amount of coulombs passed = \frac{0.142g}{0.0011178 g/C }

          amount of coulombs passed = 127.03524 C

Thus from the above conclusion we can say that amount of coulombs have passed is 127.03524 C

Learn more about Electrolysis here: brainly.com/question/16929894

#SPJ4

7 0
1 year ago
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