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denis-greek [22]
4 years ago
14

If a chemical reaction goes to completion, what can we say about the equilibrium constant for that reaction?

Chemistry
1 answer:
aniked [119]4 years ago
7 0

Answer:

A. The equilibrium constant is very large

Explanation:

The equilibrium constant value is the ratio of the concentrations of the products over the reactants. When a chemical reaction goes to completion, that means that all the reactant has turned into products. As the equilibrium constant defines, it is the ratio of the product to the reactant. So at the final stage of the chemical reaction, the equilibrium constant will be very large.

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Help me please i don’t understand the problem
brilliants [131]

Answer:

Atoms are neither created or destroyed.

Explanation:

The end result of a chemical change does not create or destroy any atoms. Matter cannot be created or destroyed, meaning the same amount of atoms exist before and after the change.

5 0
4 years ago
Read 2 more answers
How many moles are present in 45.0 g of lithium oxide (Li2O)?
BigorU [14]
Molar mass Li2O = 7.0 x 2 + 16 => 30 g/mol

1 mole Li2O ---------- 30 g
( moles Li2O ) -------- 45.0 g

moles Li2O = 45.0 x 1 / 30

moles Li2O = 45.0 / 30

= 1.50 moles of Li2O 

hope this helps!


4 0
3 years ago
Seasonal Changes of Animals in Florida Seasonal Changes of Animals in Wyoming
Lapatulllka [165]

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Explanation:

3 0
3 years ago
What process uses atomic particles of carbon and uranium to determine exact age?
Alenkinab [10]
<span>This is the process of radioactive dating. Carbon is used because it is an essential part of all living organisms and has a radioisotope that decays with a half-life of 5000 years. This ratio makes it easy to date an object using Carbon-14. Uranium is used for objects that have a much longer half-life, outside the bounds of carbon's abilities.</span>
3 0
3 years ago
Given the following equation: 4 NH3 (g)5 O2 (g) &gt;4 NO (g) + 6 H20 (I) How many moles of NH3 is required to react with 25.7 gr
Sergeu [11.5K]

Answer:

0.6425 moles of NH_3 is required to react with 25.7 grams of O_2.

Explanation:

mas of oxygen gas = 25.7 g

moles of oxygengas = \frac{25.7 g}{32 g/mol}=0.8031 mol

4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)

According to reaction given above,  5 moles of oxygen gas reacts with 4 moles of ammonia gas.

Then 0.8031 moles of oxygen gas will react with :

\frac{4}{5}\times 0.8031 mol=0.6425 mol of ammonia gas

0.6425 moles of NH_3 is required to react with 25.7 grams of O_2.

5 0
3 years ago
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