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
Too much money and dangerous
Answer:
16.05 amu
Explanation:
12.011 rounds to about 12.01 and 1.008 rounds to about 1.01 so when adding you'd do [12.01 + (1.01×4)]= 16.05
Answer:
PUT THE TOOTHPAST ON CABBAGE AND CHECK IT OUT AFTER A DAY
The best way to determine the number of atoms of arsenic in the sample will be to multiply 2.3 by Avagadro's number.
This is because Avagadro's number is the number of particles one mole of any substance has, and its value is 6.02 x 10²³
If the number of moles of a substance are known, then multiplying by Avagadro's number will give the number of particles. In this case, this is 1.38 x 10²⁴.
