To achieve the cancellation of electrons, the oxidation half-reaction needs to be multiplied by 4 while the reduction half-reaction must be multiplied by 3. Explanation: The oxidation reaction accounts for the loss of electrons, increasing the oxidation state, while the reduction implies gaining electrons, leading to a decrease in oxidation state. The respective half-reactions illustrate this, confirming that multiplying the oxidation by 4 and the reduction by 3 achieves the desired effect.
The correct equation is (C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l). A buffer system is composed of a weak acid and its corresponding salt, effectively stabilizing the pH levels within a solution. The buffer works by adjusting the concentrations of the conjugate acid and base, maintaining the pH constant.
extinction coefficient (ε) = 347 L·mol⁻¹·cm⁻¹. The chemical equation representing the reaction of chromium (Cr) with hydrochloric acid (HCl) is: 2 Cr + 6 HCl → 2 CrCl₃ + 3 H₂. To find the number of moles, we apply the formula: number of moles = mass / molar weight. For chromium, we calculate: number of moles of Cr = 0.3 × 10⁻³ (g) / 52 (g/mole), leading to number of moles of Cr = 5.77 × 10⁻⁶ moles. Examining the reaction, we observe that 2 moles of Cr yield 2 moles of CrCl₃, hence 5.77 × 10⁻⁶ moles of Cr will also produce 5.77 × 10⁻⁶ moles of CrCl₃. The molar concentration is determined by: molar concentration = number of moles / volume (L), thus molar concentration of CrCl₃ = 5.77 × 10⁻⁶ / 10 × 10⁻³, which equals 5.77 × 10⁻⁴ moles/L. To convert percent transmittance (%T) to absorbance (A), we use the equation A = 2 - log(%T). Therefore, A = 2 - log(62.5), leading to A = 0.2. The relationship defining absorbance (A) includes the extinction coefficient (ε), path length (l), and concentration (c): A = εlc, hence ε = A / lc, giving ε = 0.2 / (1 × 5.77 × 10⁻⁴), which results in ε = 0.0347 × 10⁴. Thus, the extinction coefficient is ε = 347 L·mol⁻¹·cm⁻¹.
A mixture is created by dissolving 1.43 mol of potassium chloride (KCl) in 889 g of water. The concentration of KCl works out to be 1.61 molal.
The amount of KCl is 1.43 mol
Water weighs 889 g
The molality can be calculated using the formula:
molality = moles of solute divided by kilograms of solvent
Since 1 kg equals 1000 g, 889 g is 0.889 kg.
Therefore, m = 1.43/0.889 = 1.61 molal.
