Answer:
2 H2(g) + O2(g) → 2 H2O(ℓ) ΔH = −570 kJ
Explanation:
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
brainly.com/question/14356286
If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? <span>
A. one-fourth</span>
Answer:
4.186 L
Explanation:
Using the pv=nrt equation and converting the grams of O2 into mols. After finding the number of mols by dividing 5.98 by 32 (2*the atomic weight of O) you plug that into the equation. So then you have 1*V=.186875*.08206*273 then you rearrange the equation to solve for v and get 4.186 L
