What is the percent composition of each element in S2Br

1 answer:

7 0

Get the molar mass of each element and since there are two S for each Br, make it 2x32.06 + 1x79.90 and then for S2, you would get 64.12/144.02 which is 44.52% S2 and then 55.48% Br

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Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = 0,1122 g of C ≡ 112,2mg of C

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = 0,0110 g of H ≡ 11,0mg of H

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = $\frac{112,2 mgC}{200,0mg}$ ×100 = 56,1%C

%H = $\frac{11,0 mgH}{200,0mg}$ ×100 = 5,5%H

%Cl = $\frac{55,14 mgCl}{200,0mg}$ ×100 = 27,6%Cl

%N = $\frac{21,66 mgN}{200,0mg}$ ×100 = 10,8%N

I hope it helps!

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3 years ago

URGENT!! Can someone please explain this to me?

djverab [1.8K]

Answer:

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for the balanced equation

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