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monitta
3 years ago
5

What is the volume of a 0.0355mol/L calcium hydroxide solution required to react completely with 50.0mL of 0.250mol/L aluminum s

ulfate solution
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Answer:

V₁ =  0.35 L

Explanation:

Given data:

Molarity of calcium hydroxide = 0.0355 M

Volume of aluminium sulfate = 50.0 mL ( 50/1000 = 0.05 L)

Molarity of aluminium sulfate = 0.250 M

Volume of calcium hydroxide = ?

Solution:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ =  0.250 M .  0.05 L / 0.0355 M

V₁ =  0.0125 M.L/ 0.0355 M

V₁ =  0.35 L

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You have a gas with a density of 4.8 grams per Liter and a molar mass of 75.3 grams per mole. How many liters of gas do you have
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Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos
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Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a)   Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, PF_{3} is the most ionic compound and PBr_{3} is the least ionic compound.

(b)   Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, BF_{3} is the most ionic compound and NF_{3} is the least ionic compound.

(c)   Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, TeF_{4} is the most ionic compound and BrF_{3} is the least ionic compound.

6 0
2 years ago
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