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sveta [45]
3 years ago
9

Determine the volume (liters) of 0.750 m naoh solution required to neutralize 1.50 l of 0.500 m h2so4. the neutralization reacti

on is: h2so4 (aq) + 2naoh (aq) → na2so4 (aq) + 2 h2o (l)
Chemistry
1 answer:
Elena L [17]3 years ago
8 0
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ --> Na₂SO₄  + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of H₂SO₄ moles reacted - 0.500 mol/L x 1.50 L = 0.750 mol
1 mol of H₂SO₄ reacts with 2 mol of NaOH
therefore 0.750 mol of H₂SO₄ reacts with - 2 x 0.750 = 1.50 mol of NaOH
molarity of NaOH solution is 0.750 M
there are 0.750 mol in 1 L of solution
therefore 1.50 mol in - 1.50 mol / 0.750 mol/L = 2 L

2 L of NaOH is required for neutralisation 


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3 0
3 years ago
If the half-life of C-14 is 5730 years, how much C-14 would be left from a 5000 year-old sample that started as 150 grams?
serious [3.7K]
<h3>Answer:</h3>

81.9 grams

<h3>Explanation:</h3>

From the question we are given;

  • Half-life of C-14 is 5730 years
  • Original mass of C-14 (N₀) = 150 grams
  • Time taken, t = 5000 years

We are required to determine the mass left after 5000 years

  • Using the formula;
  • N = No(1/2)^t/T​, where N is the remaining mass, N₀ is the original mass, t is the time taken and T is the half-life.

   t/T = 5000 yrs ÷ 5730 yrs

    = 0.873

N = 150 g ÷ 0.5^0.873

   = 150 g × 0.546

   = 81.9 g

Therefore, the mass of C-14 left after 5000 yrs is 81.9 g

5 0
3 years ago
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