Answer:
108.9g of Silver can be produced from 125g of Ag2S
Explanation:
The compound Ag2S shows that two atoms of Silver Ag, combined with an atom of Sulphur S to form Ag2S. We can as well say the combination ration of Silver to Sulphur is 2:1
•Now we need to calculate the molecular weight of this compound by summing up the molar masses of each element in the compound.
•Molar mass of Silver Ag= 107.9g/mol
•Molar mass of Sulphur S= 32g/mol
•Molecular weight of Ag2S= (2×107.9g/mol) + 32g/mol
•Molecular weight of Ag2S= 215.8g/mol + 32g/mol= 247.8g/mol
•From our calculations, we know that 215.8g/mol of Ag is present in 247.8g/mol of Ag2S
If 247.8g Ag2S produced 215.8g Ag
125g Ag2S will produce xg Ag
cross multiplying we have
xg= 215.8g × 125g / 247.8g
xg= 26975g/247.8
xg= 108.85g
Therefore, 108.9g of Silver can be produced from 125g of Ag2S
1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(HCl) = 1 : 1.
0,00675 mol : n(HCl) = 1 : 1.
n(HCl) = 0,00675 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0,00675 mol ÷ 0,130 mol/L.
V(HCl) = 0,0519 L = 51,92 ml.
2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(KCl) = 1 : 1.
0,00675 mol : n(KCl) = 1 : 1.
n(KCl) = 0,00675 mol.
m(KCl) = n(KCl) · M(KCl).
m(KCl) = 0,00675 mol · 74,55 g/mol.
m(KCl) = 0,503 g.
n - amount of substance.
M - molar mass.
X=0.031903 I think if you don’t know how to do this photo math would be a good thing for you
The answers are:
1. D
2. B if it is a check all that are true it is b & d
Fluorine in compounds is always assigned an oxidation number of -1
