Answer: 6.162g of Ag2SO4 could be formed
Explanation:
Given;
0.255 moles of AgNO3
0.155 moles of H2SO4
Balanced equation will be given as;
2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)
Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,
Therefore the number of moles of Ag2SO4 produced is given by,
n(Ag2SO4) = 0.255 mol of AgNO3 ×
[0.155mol H2SO4 ÷ 2 mol AgNO3] x
[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]
= 0.0198 mol of Ag2SO4.
mass = no of moles x molar mass
From literature, molar mass of Ag2SO4 = 311.799g/mol.
Thus,
Mass = 0.0198 x 311.799
= 6.162g
Therefore, 6.162g of Ag2SO4 could be formed
Answer:
C.Energy is absorbed because the products have more chemical energy than the reactants.
Explanation:
When the reaction equation is:
HgBr2(s) ↔ Hg2+(aq) + 2Br-(aq)
So Ksp expression = [Hg2+] [Br-]^2
assume the solubility S = X = 2.66 x 10^-7 M
and from the reaction equation :
we can see that [Hg2+] = X
and the [Br-] = 2 X
so by substitution in Ksp formula will can get the Ksp value:
∴ Ksp = X * (2X)^2
= 2.66 x 10^-7 * (2*2.66 x 10^-7)^2
= 7.53 x 10^-20
Answer:
The right answer is "60.56 atm".
Explanation:
As we know,
Vander wall's equation is:
⇒ 
or,
⇒ 
Here,
a = 3.59 L² atm mol⁻²
b = 0.0427 L mol⁻¹
By putting the values in the above equation, we get
⇒ 
Answer:
1.5 mol
Explanation:
Step 1: Given data
- Volume of argon gas: 33 L
- Standard temperature: 273.15 K
Step 2: Calculate the moles corresponding to 33 L of argon at standard temperature and pressure (STP)
At STP, 1 mole of argon gas occupies 22.4 L.
33 L × 1 mol/22.4 L = 1.5 mol
