1 mole Fe produces 1 mole Cu
<span> molar mass of Fe is 55.8 g / mole
answer: </span><span>55.8g</span>
Answer:
The final mass of sample is 1.3 g.
Explanation:
Given data:
Half life of H-3 = 12.32 years
Amount left for 15.0 years = 3.02 g
Final amount = ?
Solution:
First all we will calculate the decay constant.
t₁/₂ = ln² /k
t₁/₂ =12.32 years
12.32 y = ln² /k
k = ln²/12.32 y
k = 0.05626 y⁻¹
Now we will find the original amount:
ln (A°/A) = Kt
ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y
ln (3.02 g/ A) = 0.8439
3.02 g/ A = e⁰°⁸⁴³⁹
3.02 g/ A = 2.33
A = 3.02 g/ 2.33
A = 1.3 g
The final mass of sample is 1.3 g.
B because it’s got the least connections
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
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- (for some people) the stomach starts to hurt
