Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
Answer:

Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

So we solve for C2:

2. Now, since 111 mL of water is added, we compute the final volume, V3:

So, the final concentration of the 139 mL portion is:

Best regards!
They form molecules which can be in solution form if diluted in water, but some do form solutions on exposure to the atmosphere i.e they are deliquescent like pellets of sodium hydroxide