So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
that 's what I think
Answer:
Fe is limiting, and it will produce .0188 mols of Fe2O3
Explanation:
after you convert both Fe and O2 to mols by using their molar mass, you see there is less Fe than O2 so that is your limiting reactant. To find the amount of Fe2O3 you devide the limiting reactant by it's coefeciant (4) then multiply it by the products coefficant (2). Let me know if you have any questions
Answer:
D
Explanation:
Glucose is produced through a process powered by sunlight that consumes Carbon Dioxide and produces oxygen as a waste product. In order for glucose to be produced, Carbon dioxide (a compound) must be, in essence, converted into a completely different chemical: dioxygen.
Answer:
8.909*10^-4 moles
Explanation:
The mixture contains MgCl
and NaCl with a total mass of 0.4015 g. The mass of precipitate AgCl(aq) is 1.032 g and it has a molar mass of 143.32 g/mol. Therefore, the moles of the precipitate is:
n = 1.032/143.32 = 7.2007*10^-3 moles
Molar mass of NaCl = 58.44 g/mol and the molar mass of MgCl is 95.21 g/mol. Let the mass (g) of MgCl in the original mixture be 'x'. Thus:
7.2007*10^-3 = (0.4015-x)/58.44 + 2x/95.21
(7.2007*10^-3)*58.44*95.21 = 95.21(0.4015-x) + 2x(58.44)
40.06505 = 38.227 -95.21x + 116.88x
40.06505 - 38.227 = -95.21x + 116.88x
1.83805 = 21.67x
x = 1.83805/21.67 = 0.0848 g
moles of MgCl = 0.0848g/95.21 g/mol = 8.909*10^-4 moles
