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gayaneshka [121]
3 years ago
15

A 0.00100 moles sample of ca(oh)2 requires 25.00 ml of aqueous hcl for neutralization. how many moles of acid are contained in t

he 25.00 ml?
a.0.0200
d.4.00 ´ 10–5
b.0.0400
e.none of the above
c.0.002
Chemistry
1 answer:
mr Goodwill [35]3 years ago
7 0
The balanced chemical equation for the reaction is as follows;
Ca(OH)2 + 2HCl —> CaCl2 + 2H2O
The stoichiometry of Ca(OH)2 to HCl is 1:2.
Number of Ca(OH)2 moles reacted = 0.00100 mol
Therefore number of HCl moles that should react with base is double the amount of moles of base reacted.
Number of acid moles in 25.00 ml
= 0.00100 mol x2 = 0.00200 mol
Correct answer is 0.002 mol
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An atom X has three electrons in its outermost shell. Which ion will most likely be formed by X?
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Explanation:

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8 0
3 years ago
Calculate the volume in milliliters of a iron(II) bromide solution that contains of iron(II) bromide . Round your answer to sign
miv72 [106K]

This is an incomplete question, here is a complete question.

Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.

Answer  : The volume of iron(II) bromide solution is, 2.13\times 10^2mL

Explanation : Given,

Concentration of iron(II) bromide = 1.29 mo/L

Moles of iron(II) bromide = 275 mmol = 0.275 mol

conversion used : 1 mmol = 0.001 mol

Now we have to calculate the volume of iron(II) bromide.

\text{Volume of iron(II) bromide}=\frac{Moles of iron(II) bromide}}{\text{Concentration of iron(II) bromide}}

Now put all the given values in this formula, we get:

\text{Volume of iron(II) bromide}=\frac{0.275mol}{1.29mol/L}=0.213L=2.13\times 10^2mL

Thus, the volume of iron(II) bromide solution is, 2.13\times 10^2mL

5 0
3 years ago
A laboratory technician wants to determine the aspirin content of a headache pill by acid-base titration. Aspirin has a Ka of 3.
makkiz [27]

Answer:

pH = 2.97

Explanation:

Aspirin, (HC₉H₇O₄), is in equilibrium with water, thus:

HC₉H₇O₄(aq) + H₂O(l) ⇄ C₉H₇O₄⁻(aq) + H₃O⁺(aq)

Ka = 3.0x10⁻⁴ = [C₉H₇O₄⁻][H₃O⁺] / [HC₉H₇O₄]

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[C₉H₇O₄⁻] = x

[H₃O⁺] = x

Replacing in Ka formula:

3.0x10⁻⁴ = [x][x] / [0.0050-x]

1.5x10⁻⁶ - 3.0x10⁻⁴X = X²

X² + 3.0x10⁻⁴X - 1.5x10⁻⁶ = 0

Solving for x:

x = - 0.00138 → False answer. There is no negative concentrations

x = 0.00108 → Right answer

As [H₃O⁺] = x; [H₃O⁺] = 0.00108M.

pH = -log[H₃O⁺]

<em>pH = 2.97</em>

6 0
3 years ago
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