Fossil fuels and unnatural energy sources still dominate in the US, this influences many problems including global warming, climate change, and pollution. Burning fossil fuels is also a problem to human health. And unnatural resources won’t be there forever whereas, renewable sources could last a long time.
Don't understand the text, but water always freezes at 0°C as long as its water.
Answer:
Molarity: 0.111M
% (w/w): 0.666
Explanation:
The reaction of NaOH with acetic acid (CH₃COOH) is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>
As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:
0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:
4.169x10⁻³ moles of CH₃COOH.
Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:
4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>
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As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:
4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>
Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.
Thus, percent by weight is:
0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>
1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.
2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.
