Describe the cause of attraction between molecules of water

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.

To determine the mass of the lead piece, we use the following equation: Q(heat) = mC delta T, where Q equals 78.0 j, M is the mass we want to find, C is the specific heat capacity (0.130 j/g/C), and delta T shows the temperature difference, set at 9.0 c. Rearranging the formula to solve for M gives us M = Q / c delta T. By substituting in the values, we conclude that M = 78.0 j / (0.130 j/g/C * 9.0 C), calculating this gives us a mass of 66.7 g of lead.

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

• The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
• Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

The molarity of Cu²⁺ is:

Answer:

The original halide's formula is SrCl₂.

Explanation:

• The chemistry reaction's balanced equation is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.

• Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.

• The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.

• The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.

• n = mass / molar mass, thus n =  4.11 x 10⁻³ moles and mass = 0.652 g.

• The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.

• The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).

• Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.

• This identifies the atomic mass of Cl.

• Consequently, the original halide's formula is SrCl₂.

<span>Some solutions demonstrate colligative properties, which rely on the quantity of solute in a solvent. To find the elevation in boiling point, we use the formula:

</span><span>ΔT(boiling point)  = (Kb)mi

where Kb represents a constant, m is the solution's molality, and i is the van't Hoff factor.

From the provided information, we can easily determine i as follows:

</span>ΔT(boiling point)  = (Kb)mi
103.45 - 100  = (0.512)3.90i
i = 1.73 <-------van't Hoff factor