What is the H+ concentration in a solution with a pH of 1.25? ____× 10n M n= ____
2 answers:
You might be interested in
Answer:
A. 10.0 grams of ethyl butyrate would be synthesized.
B. 57.5% was the percent yield.
C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.
Explanation:
A
Moles of butanoic acid =
According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :
of ethyl butyrate
Mass of 0.08636 moles of ethyl butyrate =
0.08636 mol × 116 g/mol = 10.0 g
Theoretical yield = 10.0 g
Experimental yield = ?
Percentage yield of the reaction = 100%
Experimental yield = 10.0 g
10.0 grams of ethyl butyrate would be synthesized.
B
Theoretical yield of ethyl butyrate = 10.0 g
Experimental yield ethyl butyrate = 5.75 g
Percentage yield of the reaction = ?
57.5% was the percent yield.
C
Moles of butanoic acid =
According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :
of ethyl butyrate
Mass of 0.08636 moles of ethyl butyrate =
0.08636 mol × 116 g/mol = 10.0 g
Theoretical yield = 10.0 g
Experimental yield = ?
Percentage yield of the reaction = 78.0%
Experimental yield = 7.80 g
7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.
Answer:
The molarity of this solution is 0,09254M
Explanation:
The concentration of the Ni²⁺ solution is:
Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻
0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>
25,00 mL of this solution contain:
0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺
The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:
0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺
Thus, moles of Ni²⁺ that react with CN⁻ are:
4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺
For the reaction:
4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻
Four moles of CN⁻ react with 1 mole of Ni²⁺:
2,9450x10⁻⁴ moles of Ni²⁺ × = <em>1,178x10⁻³ moles of CN⁻</em>
As the volume of cyanide solution is 12,73mL. The molarity of this solution is:
<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>
I hope it helps!