Answer: 0.70g
The half-life of iron-53 would be 8.51 minutes. So, in 25.53 minutes would be equal to: 25.53 min/ (8.51 minutes/ half-life)= 3 half-life.
Every half-life will reduce the original weight into half. So, the final weight would be:
final weight = original weight * 1/2 ^(time)
final weight = 5.6g * (1/2)^(3 half-life)
final weight = 5.6g * 1/8= 0.7g
Given:
Moles of H2 = 0.300
Moles of I2 = 0.400
Moles of HI = 0.200
Keq = 870
To determine:
Amounts of the mixture at equilibrium
Explanation:
H2(g) + I2(g) ↔ 2HI(g)
Initial 0.3 0.4 0.2
Change -x -x +2x
Eq (0.3-x) (0.4-x) (0.2+2x)
Keq = [HI]²/[H2][I2]
870 = (0.2+2x)²/(0.3-x)(0.4-x)
x = 0.29 moles
Amounts at equilibrium:
[HI] = 0.2 + 2(0.29) = 0.78 moles
[H2] = 0.3-0.29 = 0.01 moles
[I2] = 0.4-0.29 = 0.11 moles
The formula of the product when Br2 reacts with 3L of F2 to completion and produce 2 L of the product is
2 BrF3
Br2 +3F2 = 2BrF3
1 mole of Br reacted 3 moles of f2 to form 2 moles of BrF3
The O-16 nucleus has a mass of 15.9905 amu, A proton has a mass of 1.00728 amu, a neutron has a mass of 1.008665amu, and 1amu is equivalent to 931 MeV of energy.