1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon,
then, (415.2/89.2) ×100 = 465.47 g of the substance will be required to yield 34.6 moles of carbon.
It's a because if you add them together you till get 1.40
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
Is there some kind of diagram? how is your finger pushing the coin, and where? It may be:
1)friction against a surface
2)push from the finger
3)gravity
4)air resistance behind the coin
Answer:
The coefficient of O2 is 11
Explanation:
Step 1:
The equation for the reaction:
FeS2 + O2 → SO2 + Fe2O3
Step 2:
Balancing the equation. The equation can be balance as follow:
FeS2 + O2 → SO2 + Fe2O3
There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:
2FeS2 + O2 → SO2 + Fe2O3
There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:
2FeS2 + O2 → 4SO2 + Fe2O3
Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:
2FeS2 + 11/2O2 → 4SO2 + Fe2O3
Multiply through by 2 to clear the fraction as shown below:
4FeS2 + 11O2 → 8SO2 + 2Fe2O3
Now the equation is balanced.
The coefficient of O2 is 11
