The molecular reaction between hydroiodic acid and aqueous sodium hydroxide is shown above. It is a reaction of one mole of hydroiodic acid with one mole of sodium hydroxide to yield salt and water only. It is a neutralization reaction. Hydrogen iodide dissolves in water to produce hydronium ions which agrees with Arrhenius description of acids. Hydroiodic acid is a strong acid with PKa of -9.3

3 0

3 years ago

How is molar mass used in some stoichiometric calculations?

zhenek [66]

Answer:

Explanation:

In stoichiometry, molar mass is used to convert between the mass of a substance and how many moles of a substance.

6 0

3 years ago

Please answer this correctly!

Marina CMI [18]

Answer:

International is the answer duh

4 0

1 year ago

Estimate ΔH for the reaction: C2H6(g) + Cl2(g)--> C2H5Cl(g) + HCl(g) given the following average bond energies (in kJ/mol): C

Leno4ka [110]

Explanation:

The reaction equation will be as follows.

$C_{2}H_{6}(g) + Cl_{2}(g) \rightarrow C_{2}H_{5}Cl(g) + HCl(g)$

Using bond energies, expression for calculating the value of $\Delta H$ is as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

On reactant side, from $C_{2}H_{6}$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 6

From $Cl_{2}$ ; Cl-Cl bonds = 1

On product side, from $C_{2}H_{5}Cl$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 5

C-Cl bonds = 1

From HCl; H-Cl bonds = 1

Hence, using the bond energies we will calculate the enthalpy of reaction as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

= $[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]$ = -102 kJ/mol

Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.

5 0

3 years ago