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jek_recluse [69]
3 years ago
11

What is the correct sequence of coefficients when this equation is balanced? ___cs2(l) + ___o2(g) → ___co2(g) + ___so2(g)?

Chemistry
2 answers:
Leviafan [203]3 years ago
7 0
<span> cs2(l) + 3 o2(g) → co2(g) + 2 so2(g</span>
iris [78.8K]3 years ago
4 0

Answer: the sequence is 1, 3, 1, 2.


Explanation:


1) Given equation: CS₂(l) + O₂(g) → CO₂(g) + SO₂(g)


2) To balance:


i) initially C is balanced: 1 atom in the left side, and 1 atom in the right sides


ii) Add coefficient 2 in front of SO₂ to have 2 atoms in each side.


That leads to: CS₂(l) + O₂(g) → CO₂(g) + 2SO₂(g)


iii) Count the O atoms; there are 2 atoms in the left, and 6 atoms in the right side. So, add a coefficient 3 in front of O₂ (in the left) and you get:


CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)


4) You can verigy that now the equation is balanced:


C: 1 in the left, and 1 in the right


S: 2 in the left, and 2 in the right


O: 6 in the left, and 6 in the right.


5) So, the sequence is 1, 3, 1, 2.


Remember, when the none coefficient is shown it is because it is 1.

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0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

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Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
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