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VMariaS [17]
4 years ago
9

How many moles of na+ are present in 343 ml of a 1.27 m solution of na2so4?

Chemistry
1 answer:
cricket20 [7]4 years ago
4 0
Sodium sulfate, Na₂SO₄, is a strong electrolyte. It means that it dissociates into 2 Na⁺ ions and 1 SO₄⁻ ion when dissolved in water. So, for every 1 mole of Na₂SO₄, there is an equivalent amount of 2 mol Na⁺ ions. The solution is as follows:

Moles of Na⁺ = (1.27 mol Na₂SO₄/L)*(1 L/1000 mL)*(343 mL)*(2 mol Na⁺/1 mol Na₂SO₄)
<em>Moles of Na⁺ = 0.87 mol</em>
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Answer:

The answer to your question is: letter c (96%)

Explanation:

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu)  ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula

Atomic mass = m₁(%₁) +m₂(%₂)  / 100

%₁ = x I established this is an equation

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Substituting values

114.82 = 112.8040x + 114.9039(100-x) /100    and know we expand and simplify

114.82 = 112.8040x + 11490.39 - 114.9039x  /100

11482 = 112.8040x -114.9039x +11490.39

11482 - 11490.39 = 112.8040x -114.9039x

-8.39 = -2.099x

x = 3.99

Then % of Indium-115 = 100 - 3.99 = 96

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