Muscles undergo fermentation when no carbon dioxide is available. no oxygen is available. no pyruvate is available. no water is
1 answer:
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Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
- Mg: 24.301;
- O: 15.999.
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.
Answer:
1) Fe = 69.9%
O = 31.1%
2) H = 5.19%
O = 16.5%
N = 28.9%
C = 49.5%
Explanation:
One easy way to do percent compositions is to assume you have 100g of a substance.
1) Lets say we have 100g of Fe2O3.
The total molar mass would be:
The molar mass of the Fe2 alone is:
Thus, the grams of Fe2(out of a 100) could be calculated by multiplying 100g * the molar mass ratio of Fe2 to the whole:
Which is approximately 69.9%.
We can find the amount of O3 by simply subtracting, as the rest of the compound is made of O3. Thus, the % composition of O3 is 31.1%
You can then do this same process to the next question, getting us the following:
H = 5.19%
O = 16.5%
N = 28.9%
C = 49.5%