The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL
<h3>Balanced equation </h3>
CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O
From the balanced equation above,
- The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the volume of NaOH</h3>
- Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
- Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
- Molarity of base, NaOH (Mb) = 0.242 M
- Volume of base, KOH (Vb) =?
MaVa / MbVb = nA / nB
(0.204 × 46.79) / (0.242 × Vb) = 1
Cross multiply
0.242 × Vb = 0.204 × 46.79
Divide both side by 0.242
Vb = (0.204 × 46.79) / 0.242
Vb = 39.44 mL
Thus, the volume of NaOH needed for the reaction is 39.44 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
Explanation:
Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g
1 kg = 1000 g
Molar mass of fat = M
M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]
Moles of fat =
According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give ;
of water
Mass of 148.31 moles of water ;
148.31 mol × 18 g/mol = 2,669.58 g
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
A.Weak
B.concentrated
C.dilute
D.strong
Answer: A - Weak
Reaction:
<span>HCl + NaOH ---> NaCl + H2O
</span><span>1 mole of HCl = 36,5 g
</span><span>1 mole of NaOH = 40g
</span><span>so, according to the reaction:
</span><span>1 mol HCl = 1 mol NaOH
</span>so, we need > 36,5 g HCl (<u>hydrochloric acid</u><span>)
</span><u>
answer: 36,5 g HCl (hydrochloric acid)
</u><span> ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
</span><span>next question.
</span><span>
1 mole of NaCl = 58,5 g
</span><span>1 mole of H2O = 18g
</span>
so, according to the reaction:
1 mole of HCl (36,5 g) <span>----------------- - 1 mole of NaCl (58,5 g)
</span><span>(the same for NaOH)
i
</span>1 mole of HCl<span> (36,5 g) ------------------ 1 mole of H2O (18 g)
</span>(the same for NaOH)
<span>so, this reaction is stechiometric
</span><u>
answer: 58,5 g NaCl i 18g H2O</u>