Which list includes 3 types of chemical reactions

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andrezito [222]

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5 0

3 years ago

Read 2 more answers

Dust in homes often clings to furnishings due to electrostatic interactions

Olenka [21]

This is true do to all the other particules around

3 0

3 years ago

How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCl
determine the mole ratio,
use the mole ratio to calculate the number of moles of aluminum.
use RFM of Aluminum to determine the grams required.

Moles of HCl

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0

2 years ago

The picture below shows a warm air mass caught between two cooler air masses. What is this type of front called? Plz, I really n

pentagon [3]

An occluded front forms when a warm air mass is caught between two cooler air masses.

6 0

3 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

3 0

3 years ago