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:
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D) They both look uniform (the same) throughout.
<h3>Further explanation</h3>
Pure substance can be any element or compound and is formed from one type of atom/molecule only
Meanwhile, the solution is included in a mixture consisting of 2 or more pure substance
Pure substance can be formed through a chemical process while the mixture is through a physical process
Mixture can be separated by physical processes into components of pure substance while pure substance cannot
The mixture itself consists of a homogeneous and heterogeneous solution
The mixture can be divided into a homogeneous mixture if the composition/ratio of each substance in the mixture is the same and a heterogeneous mixture if the ratio of the composition of the substances is not the same (can be varied) in each place.
Mixtures can also be divided into solutions, suspensions, and colloids based mainly on the size of the particles
Homogeneous mixture = Solution
Heterogeneous mixture = suspension, and
The mixture is located between suspension and solution = Colloid
Answer:
S+ F2 ⇒ SF
S=1
F =2
So S +F2 ......... 2SF
2S + F2 ..........2SF this is a balance equation
S=2 F=2 in left side s=2 F = 2 in rightside
Explanation:⇆
⇒
Answer:
In full volume it contain 0.12 moles.
Explanation:
Given data:
Total volume= Vt = 2.9 L
Decreased volume= Vd = 1.2 L
Number of moles of air present in decreased volume= n = 0.049 mol
Number of moles of air in total volume= n = ?
Solution:
Formula:
Vt/ Vd = n (in total volume) /n ( decreased volume)
2.9 L / 1.2 L = X / 0.049 mol
2.42 = X / 0.049 mol
X = 2.42 × 0.049
X = 0.12 mol
