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Blababa [14]
3 years ago
12

Solid iron(II) sulfide reacts with aqueous hydrochloric acid HCl to produce hydrogen sulfide gas H2S and aqueous iron(II) chlori

de . Write a balanced chemical equation for this reaction.
Chemistry
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

<h2> 2 HCl(aq) + FeS(s) --->  FeCl_{2(aq) +H_{2}S(g)</h2>

Explanation:

Here translate the words into the atomic symbol:

HCl(aq) + FeS(s) →FeCl2(aq) + H2S(g)

calculation;  

Hydrogens (H); 1 on left, 2 on right

Chloride (Cl);  1 on left, 2 on right

Iron (Fe) ; 1 on left, 1 on right

Sulfur (s);  1 on left, 1 on right

Balancing ; So, need 1 more Hydrogen on the left and 1 more Chloride on the left.  The easiest way to do this is to put a 2 in front of HCl.  

<em>2 HCl(aq) + FeS(s) --->   </em>FeCl_{2(aq) +H_{2}S(g)

Now there are two hydrogens, two chlorides, one iron, and one sulfur on the left and the same on the right.

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natima [27]

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

<h3>Rate = k×[A]</h3>

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

<h3>k = 0.213s⁻¹</h3>

6 0
3 years ago
Why is the enthalpy of formation of oxygen zero?
madreJ [45]
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4 0
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Compared to the nonmetals in Period 2, the metals in Period 2 generally have larger
konstantin123 [22]
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6 0
3 years ago
1. All ___matter______ is made of atoms that cannot be ______________________ or ____________________. 2. All _______________ of
marin [14]

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Nikitich [7]

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