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suter [353]
3 years ago
9

You titrate 25.00 ml of 0.1894 M acetic acid with 0.2006 M NaOH. How many ml of NaOH (to four significant figures) will be requi

red at the equivalence point?
Chemistry
1 answer:
Anvisha [2.4K]3 years ago
8 0

Answer:

23.60 mL NaOH

Explanation:

The reaction is CH3COOH + OH- --> CH3COO+ + H2O

Since the reaction is one-to-one, we can use M1V1 = M2V2.

M1 = 0.1894 M CH3COOH

V1 = 25.00 mL CH3COOH

M2 = 0.2006 M NaOH

V2 = ?

Solve for V2 --> V2 = M1V1/M2

V2 = (0.1894 M)(25.00 mL) / (0.2006 M) = 23.60 mL NaOH

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A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 ​ , 1.09 g h 2 ​ o, and 1.70 g n 2 ​ . if the molar mass of th
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Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) =  5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
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n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.


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