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Pepsi [2]
4 years ago
10

36 Which compound has the highest percent com­position by mass of strontium?

Chemistry
1 answer:
pav-90 [236]4 years ago
7 0
To solve this problem we must first find the masses of each individual compound. Then we can use the following equation:
(mass of strontium) / (mass of compound) * 100% = (percent mass of Strontium)

1. The mass of strontium is 87.62 and the mass of Chlorine is 35.45. There are 2 chlorine atoms in the compound, so 35.45*2 = 70.9. Add the strontium to get the mass of compound. 70.9 + 87.62 = 158.52. Now, we can use the equation. 

87.62 / 158.52 * 100% = 55.27% Strontium

2. There are 2 Iodine atoms in the compound, and the molar mass of Iodine is 126.9. The mass of the compound is 126.9 + 87.62 = 214.52. Now, we can use the  equation

87.62 / 214.52 * 100% = 40.84% Strontium

3. The molar mass of Oxygen is 16, and there is only one oxygen atom. So the mass of the compound will be 16 + 87.62 = 103.62. Now we can use the equation.

87.62 / 103.62 * 100% = 85.46% Strontium

4. Finally, the molar mass of sulfur is 32.06, and there is only one sulfur atom in the compound, so the mass will be 32.06 + 87.62 = 119.68. Now we can use the equation

87.62 / 119.68 * 100% = 73.21% Strontium

Therefore, the answer is 3 because it has the highest percentage of Strontium than the other compounds. Hope this helped!! :D
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Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

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The given chemical reaction is,

IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)

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First balance the main element in the reaction.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

Now balance oxygen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-+3H_2O

Now balance hydrogen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-+6H^+\rightarrow I^-+3H_2O

Now balance the charge.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}+2e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : 2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

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IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

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