Answer:
780 g
Explanation:
Step 1: Given data
Moles of sodium phosphate: 4.76 moles
Step 2: Calculate the mass corresponding to 4.76 moles of sodium phosphate
To convert moles to mass, we need a conversion factor. In this case, we will use the molar mass of sodium phosphate, which is 163.94 g/mol.
4.76 mol × 163.94 g/mol = 780 g
A glucose molecule is completely broken down to carbon dioxide and water in glycolysis and the citric acid cycle, but together these two<span> processes yield only a few molecules of ATP.</span>
Answer:
16.5 atm
Explanation:
<em>A mixture of He, N₂, and Ar has a pressure of 24.1 atm at 28.0 °C. If the partial pressure of He is 3013 torr and that of Ar is 2737 mm Hg, what is the partial pressure of N₂?</em>
The total pressure of a gaseous mixture is equal to the sum of the partial pressures.
P = pHe + pN₂ + pAr
pN₂ = P - pHe - pAr [1]
We need to express pHe and pAr in atm.
From [1],
pN₂ = 24.1 atm - 3.96 atm - 3.60 atm = 16.5 atm
You need to add the last substance to the products side(the right sode of the arrow). You have hydrogen and oxygen - water.
You get: BrO3 + N2H4 -> Br2 + N2 + H2O
# of Br: 1x1 = 1 # of Br: 2x1 = 2
O: 3x1 = 3 O: 1x1 = 1
N: 2x1 = 2 B N: 2x1 = 2
H: 4x1 = 4. H: 2x1 = 2
Br:
Multiply the reactant (left) side by 2 to balance.
O:
You've just multiplied the reactant oxygen by 2 so now the reactant side equals 6. Multiply the product (right) side by six as well.
H:
The product side is now equal to 12. Multiply the reactant side by 3 to balance.
N:
Now you have to balance N because the reactant side has been risen. So multiply the product side by three as well.
You end up with the complete and balanced equation:
2BrO3 + 3N2H4 -> Br2 + 3N2 + 6H2O
Answer:
See below.
Explanation:
The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.
The balanced equation is:
2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)
From the equation, using atomic masses:
228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O
So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29
= 42.43 g of CO2.
Amount of water = 324.3 * 13.759) / 228.29
= 19.55 g of H2O.