Answer: 24.13 g Cu
Explanation:
<u>Given for this question:</u>
M of CuO = 30 g
m of CuO = 79.5 g/mol
Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol
= 0.38 mol
The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:
CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)
The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side
4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)
From the stoichiometry of the balanced equation:
4 moles of CuO gives 4 moles of Cu
1 mole of CuO gives 1 mol of Cu
0.38 mol of CuO gives 0.38 mol of Cu
Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu
= 0.38 × 63.5 g
= 24.13 grams
Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane
the answer is c.50atm
Explanation:
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Answer:
2KClO3 —> 2KCl + 3O2
The coefficients are 2, 2, 3
Explanation:
From the question given above, we obtained the following equation:
KClO3 —> 2KCl + 3O2
The above equation can be balance as follow:
There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:
2KClO3 —> 2KCl + 3O2
Now, the equation is balanced.
Thus, the coefficients are 2, 2, 3
From the periodic table:
mass of carbon = 12 grams/mole
mass of hydrogen = 1 gram/mole
mass of oxygen = 16 grams/mole
The given compound C₆H₁₂O₆ has 6 moles of carbon, 12 moles of hydrogen and 6 moles of oxygen.
Therefore, the molar mass of the given compound will be calculated as follows:
molar mass = 6(12) + 12(1) + 6(16) = 180 grams/mole
Answer:
a)
b)
hours
Explanation:
Given
Surface Area (S) of 250 grams of powder sample 
square meter.
square centimeter
grams/minute
mg per second
gram per cubic meter 
mg/ml
a)
cm /sec
b) Time taken
Time taken to sink
Time in hours
hours
