Chemistry???

Question

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Answer 1

Step 1:

(I)

The final volume of is 100 mL = 0.1 L.

Molar Concentration of is 0.200 M

The amount of required in moles is

Molar Mass of is 84.007 g/mol.

Amount of in  grams is

Amount of required is 1.6801 grams.

Solution:

(I) Amount of required to prepare 100 mL of is 1.6801 grams.

Answer 2

Step 1:

(II)

The final volume of is 100 mL = 0.1 L.

Molar Concentration of is 0.200 M

The amount of required in moles is

Molar Mass of is 120.366 g/mol.

Amount of in  grams is

Amount of is 2.40 grams.

Solution:

Amount of required to prepare 100 mL of is 2.40 grams.

Answer 3

Step 1:

(III)

Initial Molar Concentration or of  is 0.0200 M

The final volume of or is 50 mL = 0.05 L.

Final Molar Concentration of or is 0.0100 M

Dilution of Solution formula: ,

where is initial molar concentration.

is final molar concentration

is initial volume.

is final volume.

Initial volume is .

Solution:

Initial volume of or is .

Answer 4

Step 1:

(IV)

Initial Molar Concentration of  is 6.0 M

Final Molar Concentration of is 3.0 M

The final volume of is 100.0 mL.

Dilution of Solution formula: ,

where is initial molar concentration.

is final molar concentration

is initial volume.

is final volume.

Initial volume of is .

Solution:

Initial volume of is .

Answer 5

Step 1:

(V)

Final Molar Concentration of is 1.0 M

The final volume of is 250.0 mL.

The initial volume of is 50.0 mL.

Dilution of Solution formula: ,

where is initial molar concentration.

is final molar concentration

is initial volume.

is final volume.

Initial molar concentration of is .

Solution:

Initial molar concentration of is .

Answer 6

Step 1:

(VI)

The final volume of is 100 mL = 0.1 L.

Molar Concentration of is 0.200 M

The amount of required in moles is

Molar Mass of is 36.46 g/mol.

Amount of in  grams is

Amount of required is 0.7292 grams.

Solution:

Amount of required to prepare 100 mL of is 0.7292 grams.