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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.