How much concentrated 15.54 M sulfuric acid is need to prepare 6.1 mL of a 2.28 M solution?

0.895 mL

Explanation:

Using the formula;

C1V1 = C2V2

Where;

C1 = initial concentration (M)

C2 = final concentration (M)

V1 = initial volume (mL)

V2 = final volume (mL)

According to the information provided in this question;

C1 = 15.54M

V1 = ?

C2 = 2.28M

V2 = 6.1 mL

Using C1V1 = C2V2

V1 = C2V2 ÷ C1

V1 = (2.28 × 6.1) ÷ 15.54

V1 = 13.908 ÷ 15.54

V1 = 0.895 mL

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a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.