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Anit [1.1K]
3 years ago
8

According to Le Chatelier’s principle, what happens when the concentration of reactants is doubled in a chemical reaction that w

as in equilibrium?
Chemistry
2 answers:
Travka [436]3 years ago
6 0

Answer:

A) The equilibrium shifts to the right to favor the formation of products.

Explanation:

Usimov [2.4K]3 years ago
3 0
If more reactant is added, the equation will shift to the right in order to make more product (which will increase the products)
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Define the term. Theory
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Answer:

a supposition or a system of ideas intended to explain something, especially one based on general principles independent of the thing to be explained.

(from goo gle)

A explanation for something that doesn't seem to be answered completely, usually very implausible.

Explanation:

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3 years ago
Massive amounts of energy are contained in an atom's: Select one:
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Nucleus.
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Worth many points (timed test)
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Answer:

2.335 J/g*degrees celcius

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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
Fill in the blanks to complete the statements about the gases in Earth’s atmosphere. Earth’s atmosphere is mostly nitrogen, whic
vivado [14]

Answer:

1) Oxygen

2) water vapor

3) carbon dioxide

8 0
3 years ago
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