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Nataliya [291]
3 years ago
13

How many grams of CaOH2 are in this solution? The molar mass of CaOH2 is 58.093 g/mol. (0.0787 mol CaOH2) (58.093 g/mol) = g CaO

H2
Chemistry
1 answer:
Alenkinab [10]3 years ago
4 0

Answer:

mass is 4.57 g

Explanation:

Ca(OH)₂ solution consists of Ca(OH)₂ solute molecules and solvent and we are asked to find the mass of Ca(OH)₂ in the solution

number of moles of Ca(OH)₂ is - 0.0787 mol

molar mass of Ca(OH)₂ is - 58.093 g/mol

we can use the following equation

number of moles = mass of Ca(OH)₂ / molar mass of Ca(OH)₂

rearranging the equation

mass of Ca(OH)₂ = number of moles x molar mass

 mass = 0.0787 mol x 58.093 g/mol = 4.57 g

mass of Ca(OH)₂ is 4.57 g

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Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

I 0.20 0 0

C (−x) (+x) (+x)

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You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to

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3 years ago
A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

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And <u>converting into L </u>becomes:

  • 295 mL * \frac{1 L}{1000mL} = 0.295 L

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3 years ago
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