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zhenek [66]
3 years ago
13

Classify the outcomes based on whether they are caused by adding or removing a reactant from a chemical reaction. The rate of th

e forward reaction increases. The rate of the reverse reaction increases. The concentration of products increases. The concentration of products decreases.
Chemistry
1 answer:
Mama L [17]3 years ago
3 0

Explanation: To study the outcomes, we will apply Le-Chatelier's principle.

Le-Chatelier's principle states that if there is any disturbance in the conditions of the dynamic equilibrium, the position of equilibrium will counteract the change.

1) The rate of the forward reaction increases.

This will happen when we add the reactant to a chemical reaction. According to Le-Chatelier's principle, by increasing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, forward reaction is favored.

2) The rate of the revere reaction increases.

This will happen when we remove the reactant from a chemical reaction. According to Le-Chatelier's principle, by removing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, reverse reaction is favored.

3) The concentration of product increases.

This will happen when we add reactants to a chemical reaction. According to  Le-Chatelier's principle, when we increase the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the forward direction which means that the concentration of product will increase.

4) The concentration of products decreases.

This will happen when we remove reactants from a chemical reaction. According to  Le-Chatelier's principle, when we decrease the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the reverse direction which means that the concentration of reactants will increase or concentration of products will decrease.

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What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?
Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

6 0
2 years ago
What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
Inga [223]

Answer:

90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

It is known that,

1 mole = 22.414 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

There are 6 moles of water = 6×22.414 = 134.5 L

Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

8 0
3 years ago
How many atoms are in 5 grams of H2O?
mojhsa [17]

Hello!

We know that by the Law of Avogrado, for each mole of substance we have 6.02 * 10²³ atoms, if:

The molar mass of water (H2O)

H = 2 * (1u) = 2u

O = 1 * (16u) = 16u

---------------------------

The molar mass of H2O = 2 + 16 = 18 g / mol

If:

1 mol we have 6.02 * 10²³ atoms

1 mole of H2O we have 18 g

Then we have:

18 g ------------- 6.02 * 10²³ atoms

5 g -------------- x

\dfrac{18}{5} = \dfrac{6.02*10^{23}}{x}

18*x = 5*6.02*10^{23}

18\:x = 3.01*10^{24}

x = \dfrac{3.01*10^{24}}{18}

\boxed{\boxed{x \approx 1.672*10^{23}\:atoms}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

6 0
3 years ago
Read 2 more answers
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
When most fuels burn, water and carbon dioxide are the two main products. Why can´t you say that water and carbon dioxide are pr
STatiana [176]
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Fe(s) + O2(g) => FeO2(s)
7 0
3 years ago
Read 2 more answers
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