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sleet_krkn [62]
3 years ago
12

Please this is urgent! I need urgent help with number 8! Please help my final is tomorrow!

Chemistry
1 answer:
KatRina [158]3 years ago
3 0
B is the answer to this question
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The Following questions pertain to a 2.2M solution of hydrocyanic acid at 25°C. pKa = 9.21 at 25°C. Find the concentrations of a
horsena [70]
1) Chemical reaction: HCN + H₂O → CN⁻ + H₃O⁺.
c(HCN) = 2,2 M = 2,2 mol/L.
pKa(HCN) = 9,21.
Ka = 6,16·10⁻¹⁰.
[CN⁻] = [H₃O⁺] = x.
[HCN<span>] = 2,2 M - x.
</span>Ka = [CN⁻] · [H₃O⁺] / [HCN].
6,16·10⁻¹⁰ = x² / 2,2 M -x.
Solve quadratic equation: [CN⁻] = [H₃O⁺] = 0,0000346 M.
[HCN] = 2,2 M - 0,0000346 M = 2,199 M.

2) pH = - log[H₃O⁺].
pH = -log( 0,0000346 M).
pH = 4,46.
Hydrocyanic acid and hydronium ion (H₃O⁺) are acids. Cyanide anion (CN⁻) is the strongest base in the system, cyanide anion accept protons in chemical reaction.
pKb = pKw - pKa.
pKb = 14 - 9,21 = 4,79.

6 0
3 years ago
CH4 + 2O2 —&gt; 2H2O + CO2
Dovator [93]

Answer:  20 mol CH4 x 1CO2 / 1CH4 = 20mol  CO2

Answer: 20mol CO2

Explanation:

4 0
3 years ago
What Happens When extra Solute is added to unsaturated ,saturated and supersaturated solution​
Eduardwww [97]

Answer:

If More solute is added and it does not dissolve,then the original solution was saturated. if the added solute dissolves,then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.

3 0
3 years ago
2. A solution is made by adding 1.23 mol of KCl to 1000.0 g of water. Assume that the
ladessa [460]

Answer:

a. 74.55 g/mol.

b. 91.70 g.

c. 8.40%.

d. 1.23 mol/L.

Explanation:

<em>a. Calculate the formula weight of KCl.</em>

∵ Formula weight of KCl = atomic weight of K + atomic weight of Cl

atomic weight of K = 39.098 g/mol, atomic weight of Cl = 35.45 g/mol.

∴ Formula weight of KCl = atomic weight of K + atomic weight of Cl = 39.098 g/mol + 35.45 g/mol = 74.548 g/mol ≅ 74.55 g/mol.

<em>b. Calculate the mass of KCl in grams.</em>

  • we can use the relation:

<em>no. of moles (n) = mass/molar mass.</em>

∴ mass of KCl = n*molar mass = (1.23 mol)*(74.55 g/mol) = 91.69 g ≅ 91.70 g.

<em>c. Calculate the percent by mass of KCl in this solution.</em>

The mass % of KCl = (mass of KCl/mass of the solution) * 100.

mass of KCl = 91.70 g,

mass of the solution = 1000.0 g of water + 91.70 g of KCl = 1091.70 g.

∴ The mass % of KCl = (91.70 g/1091.70 g)*100 = 8.399% ≅ 8.40%.

<em>d. Calculate the molarity of the solution.</em>

Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(Volume of the solution (L))</em>

no. of moles of KCl = 1.23 mol,

Volume of the solution = mass of water / density of water = (1000.0 g)/(1.00 g/mL) = 1000.0 mL = 1.0 L.

M = (1.23 mol)/(1.0 L) = 1.23 mol/L.

3 0
4 years ago
How do the 5 branches of chemistry overlap?
mestny [16]

Answer:

Chemistry, Biology, Medicine, Physics, and Geology

7 0
3 years ago
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