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il63 [147K]
3 years ago
5

An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution to give 0.325 g of AgCl. What is the concentration, in moles per li

ter, of the AlCl3 solution? Must show your work on scratch paper to receive credit. AlCl3(aq) + 3 AgNO3(aq) → 3 AgCl(s) + Al(NO3)3(aq)
Chemistry
1 answer:
bezimeni [28]3 years ago
5 0

Answer:

4.07x10⁻³M AlCl₃.

Explanation:

Based on the reaction:

AlCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Al(NO₃)₃(aq)

<em>That means 1 mole of AlCl₃ reacts with 3 moles of AgNO₃ to produce 3 moles of AgCl.</em>

As 0.325g of AgCl are produced. Moles of AgCl are (Molar mass AgCl: 143.32g/mol):

0.325g AgCl ₓ ( 1 mol / 143.32g) = 2.27x10⁻³ moles of AgCl

As 3 moles of AgCl are produced from 1 mole of AlCl₃, moles of AlCl₃ that produce 2.27x10⁻³ moles of AgCl are:

2.27x10⁻³ moles of AgCl ₓ (1 mole AlCl₃ / 3 moles AgCl) =

<em>7.56x10⁻⁴ moles AlCl₃</em>

As volume of the AlCl₃ solution that reacts is 185.5mL = 0.1855L, molar concentration of the solution is:

7.56x10⁻⁴ moles AlCl₃ / 0.1855L =

<h3>4.07x10⁻³M AlCl₃</h3>
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rosijanka [135]

Answer:

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Explanation:

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=> 6 + X₂ = 6.94004

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2 years ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

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3 years ago
Please solve quickly
slavikrds [6]

Answer:

Explanation:

mass of one virus = 9.0 x 10⁻¹² mg

mass of one mole = 6.02 x 10²³ x mass of one virus

= 6.02 x 10²³ x 9.0 x 10⁻¹²

= 54.18 x 10¹¹ mg

= 54 x 10⁸ g .

= 54 x 10⁵ kg .

b )

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n x 54 x 10⁵ = 3 x 10⁷

n = 5.5555  

rounding off to 2 significant figure

5.6 moles Ans .

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