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
Emitted/given off. exo is a way to remember that is is going away. think of a flame. the heat is emitted off onto you.
Answer:
the atomic particles in the necluse are called ions which are positive and negive charged atoms
Explanation:
Hey there!
Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance PO₄.
1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance H.
6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Balance Ca.
1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Our final balanced equation:
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Hope this helps!
Answer:
Explanation:
Hello,
In this case, the combustion of methane is shown below:
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
Best regards.
