I believe the answer is Na because it should be sodium which is Na.
Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?
Answer:
The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass
Explanation:
2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.
We know that density is equal to ratio of mass to volume.
Here density of solution is 1.01 g/mL.
So mass of 1000 mL solution = (
) g = 1010 g
molar mass of NaCl = 58.44 g/mol
So mass of 2.23 moles of NaCl = (
) g = 130.3 g
% by mass is ratio of mass of solute to mass of solution and then multiplied by 100.
Here solute is NaCl.
So % by mass of 2.23 M aqueous solution of NaCl =
% = 12.90%
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
A: 6.37gNaCl
B: 1 mol NaCl
C: 58.5 g NaCl
These answers are correct on e2020 I just got them right.