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Afina-wow [57]
2 years ago
10

What is the molecular formula of Ferric Oxide with criss-cross method ​

Chemistry
1 answer:
Ksenya-84 [330]2 years ago
6 0

Answer: Fe2O3

Explanation: By applying crisscross method, the formula will be Fe2O3.

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What is the power produced by a light bulb with a voltage of 120 volts and a current of 0.045 amps? PLEASE HELP I HAVE 5 MINUTES
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The answer is 5.4! I found it on a an online calculator! 
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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

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Zainab is in science class, and he is studying an organism that reproduces sexually. The genetic material of the organism
Tanzania [10]

Answer:

The genetic material of the organism will be DNA.

Explanation:

Sexual reproduction can be described as a type of reproduction in which offsprings with genetic diversity are produced. Sexual reproduction occurs by the process of meiosis.

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The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp
pav-90 [236]

<u>Answer:</u> The initial concentration of hydrogen peroxide at the given temperature is 0.399 M

<u>Explanation:</u>

Decomposition of hydrogen peroxide is following first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant = 2.54\times 10^{-4}s^{-1}

t = time taken for decay process = 855 s

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.321 M

Putting values in above equation, we get:

2.54\times 10^{-4}s^{-1}=\frac{2.303}{855s}\log \frac{[A_o]}{0.321}

[A_o]=0.399M

Hence, the initial concentration of hydrogen peroxide at the given temperature is 0.399 M

7 0
3 years ago
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