Consider the decomposition of the compound C5H6O3 as follows below. C5H6O3(g) → C2H6(g) + 3 CO(g) When a 5.63-g sample of pure C

The question lacks completeness; the full question is:

Determine the theoretical yield:

When excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide, sodium nitrate and copper(II) sulfide precipitate. For this reaction, 469 grams of copper(II) nitrate was combined with 156 grams of sodium sulfide yielding 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Calculating moles of copper(II) nitrate =

Moles of sodium sulfide =

Based on the reaction, 1 mole of copper(II) nitrate reacts with 1 mole of sodium sulfide.

Thus, 2 moles of sodium sulfide will react with:

of copper(II) nitrate

Since sodium sulfide is in limiting quantities, the amount of sodium nitrate produced will depend on the moles of sodium sulfide available.

According to the reaction, 1 mole of sodium sulfide generates 2 moles of sodium nitrate; thus, 2 moles of sodium sulfide will yield:

sodium nitrate

The total mass of 4 moles of sodium nitrate is:

85 g/mol × 4 mol = 340 g

The theoretical yield of sodium nitrate amounts to 340 g.

The theoretical yield of sodium nitrate is 340 grams.

Answer:

The molality of the solution is 1.08 m.

Explanation:

First, determine the mass of the solvent.

A 13% solution by mass indicates that 13 grams are found in every 100 grams of solution.

Thus, solution mass = solute mass + solvent mass

100 g = 13 g + solvent mass

Therefore, solvent mass = 100 g - 13 g → 87 g

Next, we calculate the moles of solute (mass / molar mass):

13 g / 138.2 g/mol = 0.094 moles

Finally, to find the molality, which is the moles of solute per 1 kg of solvent (mol/kg), we convert the solvent mass to kg:

87 g. 1 kg / 1000 g = 0.087 kg

Then, molality → 0.094 mol / 0.087 kg = 1.08 m

Answer:

pH = 8.0

Explanation:

Initially, we need to determine the moles of NaOH.

Consider the balanced reaction.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio among C₂H₄O₃, NaOH, and C₂H₃O₃Na is 1: 1: 1. Therefore, if 7.2 × 10⁻⁴ moles of NaOH completely react with 7.2 × 10⁻⁴ moles of C₂H₄O₃, they will yield 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

To find the concentration of C₂H₃O₃Na, we note that:

C₂H₃O₃Na dissociates as follows:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

The anion C₂H₃O₃⁻ originates from a weak acid, which thus goes through basic hydrolysis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

Given that the pKa for C₂H₄O₃ is 3.9, we can derive pKb for C₂H₃O₃⁻ using this relation:

pKa + pKb = 14

pKb = 14 - 3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can compute [OH⁻] through the equation:

[OH⁻] = √(Kb.Cb)               where Cb indicates the base's initial concentration

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

We can then find pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

Initially we need to perform the conversion:145 pm = 145 * 10^(-12) m; 36 cm = 360 mm = 360 * 10^(-3) m. Then, we can calculate: 360 * 10^(-3) m / 145 * 10^(-12) m = 360 * 10^(-3) * 10^(12) / 145 = 2.482758621 * 10^(9) or:2,482,758,621 atoms.

The following is the balanced half-reaction. A reduction reaction is characterized by an atom gaining electrons, resulting in a decrease in its oxidation number. During the reduction of bromine gas to bromide ions, two electrons are transferred. The corresponding chemical equation for this transition is provided above.