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aev [14]
3 years ago
8

In this redox reaction, identify the reducing agent. 2Li + F2 2LiF

Chemistry
2 answers:
mel-nik [20]3 years ago
8 0
As what we see in the chemical reaction that is given, we can conclude that the Lithium must be the reducing agent since it is the one that is oxidised. In addition to that, a redox reaction or simply known as the oxidation-reduction reaction usually operates when electrons get transferred between species.
kvv77 [185]3 years ago
6 0
I simply refer to the correct answer:
brainly.com/question/1621662

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Difference between element and radical​
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2 years ago
Consider the reversible reaction, A+B⇌C+D. If the concentration of product D is increased, the rate of the reverse reaction woul
natali 33 [55]

Answer:

If the concentration of product D is increased, the rate of the reverse reaction would increase.

Explanation:

Chemical reaction:

A + B ⇄ C + D

In given condition the equilibrium is disturb by increasing the concentration of product.

When the concentration of product D is increased the system will proceed in backward direction in order to regain the equilibrium. Because when the product concentration is high it means reaction is not on equilibrium state the reaction will proceed backward direction to regain the equilibrium state.

According to the Le- Chatelier principle,

At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.

The equilibrium can be disturb,

By changing the concentration

By changing the volume

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4 0
3 years ago
How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield
Katyanochek1 [597]

Answer:

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

x= 2× 0.057= 0.114 moles of iron III chloride

But

Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL

4 0
3 years ago
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