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Gre4nikov [31]
3 years ago
8

A 25.0 mL solution of a monoprotic weak acid is titrated with 0.10 M NaOH. The equivalence point is reached after 38.52 mL of Na

OH has been added. What was the initial concentration of the weak acid
Chemistry
1 answer:
Sveta_85 [38]3 years ago
8 0

Answer:

0.154 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 25.0mL

M₂ = 0.10 M

V₂ = 38.52 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂  

Putting the respective values ,

M₁ * 25.0mL =  0.10 M * 38.52 mL

M₁ = 0.154 M

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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
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Answer:

65.08 g.

Explanation:

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<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
Sorry if it is blurry
Zina [86]
I really cant read it sorry i tried
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3 years ago
Read 2 more answers
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