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True [87]
3 years ago
15

Calculate each of the following quantities. (a) Mass (g) of solute in 175.4 mL of 0.267 M calcium acetate WebAssign will check y

our answer for the correct number of significant figures. g (b) Molarity of 597 mL solution containing 21.1 g of potassium iodide WebAssign will check your answer for the correct number of significant figures. M (c) Amount (mol) of solute in 145.6 L of 0.703 M sodium cyanide WebAssign will check your answer for the correct number of significant figures. mol
Chemistry
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

a) 7.40 grams

b) 0.213 M

c) 102 moles

Explanation:

(a) Mass (g) of solute in 175.4 mL of 0.267 M calcium acetate

Step 1: Data given

Volume = 175.4 mL = 0.1754 L

Molarity = 0.267 M

Molar mass = 158.17 g/mol

Step 2: Calculate moles

Moles = molarity * volume

Moles = 0.267 M * 0.1754 L

Moles = 0.0468 moles

Step 3: Calculate mass

Mass = 0.0468 moles * 158.17 g/mol

<u>Mass = 7.40 grams</u>

b) Molarity of 597 mL solution containing 21.1 g of potassium iodide

Step 1: Data given

Volume = 597 mL = 0.597 L

Mass = 21.1 grams

Molar mass KI = 166.0 g/mol

Step 2: Calculate moles KI

Moles KI = 21.1 grams / 166.0 g/mol

Moles KI = 0.127 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.127 moles / 0.597 L

Molarity = 0.213 M

(c) Amount (mol) of solute in 145.6 L of 0.703 M sodium cyanide

Step 1: Data given

Volume = 145.6 L

Molarity = 0.703 M

Step 2: Calculate moles

moles = molarity * volume

Moles = 0.703 M * 145.6 L

Moles = 102 moles

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Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

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