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Lana71 [14]
3 years ago
15

What is the pOH of a solution with [OH'] = 9,0 x 10-7?

Chemistry
1 answer:
bearhunter [10]3 years ago
6 0

Answer: D (6.04) is the best answer

Explanation:

pOH = - log [OH⁻]

pOH = - log (9.0 x 10⁻⁷)

pOH = 6.0457

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Does anyone know the answer to these?
Rashid [163]

Answer:

The answer to your question is below

Explanation:

Data

mass of CaCO₃ = 155 g

mass of HCl = 250 g

mass of CaCl₂ = 142 g

reactants = CaCO₃ + HCl

products = CaCl₂ + CO₂ + H₂O

1.- Balanced chemical reaction

             CaCO₃ + 2HCl   ⇒    CaCl₂ + CO₂ + H₂O

2.- Limiting reactant

molar mass of CaCO₃ = 40 + 12 + 48 = 100 g

molar mass of HCl = 2[1 + 35.5 ] = 73 g

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experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃

3.-

Calculate the molar mass of CaCl₂

CaCl₂ = 40 + 71 = 111 g

          100 g of CaCO₃ ------------------ 111 g of CaCl₂

           155 g of CaCO₃ ----------------- x

               x = (155 x 111) / 100

               x = 17205 / 100

              x = 172.05 g of CaCl₂

4.- percent yield

Percent yield = 142 / 172.05 x 100 = 82.5 %

5.- Excess reactant

    100 g of CaCO₃  -------------------- 73 g of HCl

     155 g of caCO₃ ------------------- x

           x = (155 x 73)/100

           x = 133.15 g

Mass of HCl = 250 - 133.15

                    = 136.9 g

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3 years ago
Use the following information to answer Q1-Q3
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