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kiruha [24]
4 years ago
9

Chem help due in 2 hours. please help. problem 14.

Chemistry
1 answer:
Murljashka [212]4 years ago
6 0
14 )

a) 1 <span>HBr + 1 NaHCO</span>₃<span> = 1 CO</span>₂<span> + 1 H</span>₂<span>O + 1 NaBr

b) </span><span>2 HNO</span>₃<span> + 1 K</span>₂SO₃<span> = 1 H</span>₂<span>O + 2 KNO</span>₃<span> + 1 SO</span>₂

c) 2 HI<span> + 1 </span>Na₂S<span> = 1 </span>H₂S<span> + </span>2 <span>NaI

d) 1 </span><span>(NH</span>₄<span>)</span>₂<span>SO</span>₄<span> + 1 Ca(OH)</span>₂<span> = 1 CaSO</span>₄<span> + 2 H</span>₂<span>O + 2 NH</span>₃
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A 34.53 ml sample of a solution of sulfuric acid, h2s04, reacts with 27.86 ml of 0.08964 m naoh solution. calculate the molarity
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The balanced equation between NaOH and H₂SO₄ is as follows
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stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol 
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number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol 
molarity of H₂SO₄ is 0.03617 M
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