At 80 ∘C, Kc=1.87×10^−3 for the reaction PH3BCl3(s)⇌PH3(g)+BCl3(g)

Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

The molar mass of Zn is 65 g/mole

and,

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

According to the balanced reaction we find that

1 mole of reacts with 2 moles of

Thus, 0.0206 moles of react with moles of

This leads us to determine that is the excess reagent because the moles provided exceed the required moles, while is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat =

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.

Answer:

1. 192.0 g/mol.

2. 84.0 g/mol.

Explanation:

• The molar mass refers to the weight of all atoms combined in a molecule measured in grams per mole.

• To find a molecule's molar mass, we begin by looking up the atomic weights of the relevant elements from the periodic table. Next, we tally the atoms present and multiply that by their respective atomic weights.

1. Molar mass of citric acid (C₆H₈O₇):

Molar mass of C₆H₈O₇ = 6(atomic mass of C) + 8(atomic mass of H) + 7(atomic mass of O) = 6(12.0 g/mol) + 8(1.0 g/mol) + 7(16.0 g/mol) = 192.0 g/mol.

2. Molar mass of baking soda (NaHCO₃):

Molar mass of NaHCO₃ = (atomic mass of Na) + (atomic mass of H) + (atomic mass of C) + 3(atomic mass of O) = (23.0 g/mol) + (1.0 g/mol) + (12.0 g/mol) + 3(16.0 g/mol) = 84.0 g/mol.

Although multiple values are given, our focus is on HCl.

<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>

moles = 0.215 L × 0.300 M

<span>moles = 0.0645 moles of HCl</span>

Response:

Clarification:

Hello,

In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

Best wishes.