A 10 gram sample of water is heated to 105 ℃ and is mixed with a 25 gram sample of water cooled to 25℃ . What is the final tempe
1 answer:
You might be interested in
<u>Answer:</u> The molarity of chloride anions in the solution is 0.01344 M
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
Given mass of ferric chloride = 72.7 mg = 0.0727 g (Conversion factor: 1 g = 1000 mg)
Molar mass of ferric chloride = 162.2 g/mol
Volume of solution = 100 mL
Putting values in above equation, we get:
1 mole of ferric chloride contains 1 mole of ions and 3 moles of
ions
Moles of = (3 × 0.00448) = 0.01344 M
Hence, the molarity of chloride anions in the solution is 0.01344 M
Find grams of C and H, using molar masses:
12.0 g C x (1.900 g CO2 / 44.0 g CO2) = 0.5182 grams C
2.02 g H x (1.070 g H2O /18 g H2O) = 0.1201 grams H
Since all the C and H in CO2 and H2O will come from the sample except oxygen, because we need to provide oxygen for sample to burn.
0.9835 g sample - 0.5182 g O - 0.1201 g H = 0.3452 grams O
find moles, using molar masses:
0.5182 grams C / 12.0 g/mol C = 0.0432 moles C
0.1201 grams H / 1.01 g/mol H = 0.119 moles H
0.3452 grams O / 16.0 g/mol O = 0.0216 moles O
0.0216 is lesser, so use it to normalize to find the ratio.
find ratios:
0.0432 moles C / 0.0216 = 2 moles C
0.119 moles H / 0.0216 = 5.5 moles H
0.0216 moles O / 0.0216 = 1 mole O
so, the ratio is C2 H{5.5} O1
double the ration to eliminate decimals.
gives your answer is
12.0 g C x (1.900 g CO2 / 44.0 g CO2) = 0.5182 grams C
2.02 g H x (1.070 g H2O /18 g H2O) = 0.1201 grams H
Since all the C and H in CO2 and H2O will come from the sample except oxygen, because we need to provide oxygen for sample to burn.
0.9835 g sample - 0.5182 g O - 0.1201 g H = 0.3452 grams O
find moles, using molar masses:
0.5182 grams C / 12.0 g/mol C = 0.0432 moles C
0.1201 grams H / 1.01 g/mol H = 0.119 moles H
0.3452 grams O / 16.0 g/mol O = 0.0216 moles O
0.0216 is lesser, so use it to normalize to find the ratio.
find ratios:
0.0432 moles C / 0.0216 = 2 moles C
0.119 moles H / 0.0216 = 5.5 moles H
0.0216 moles O / 0.0216 = 1 mole O
so, the ratio is C2 H{5.5} O1
double the ration to eliminate decimals.
gives your answer is
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
By rearrangement:
Multiplying through by (-) and solving the quadratic equation:
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
= 10.71%