We are given with the total mass of fertilizer which is 14.1 oz. This is equivalent to 399.73 grams. 15% of the total is the amount of nitrogen. Thus, the nitrogen amount is 59.96 grams.
Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
Answer:
2 moles
Explanation:
In one mole of O2 there are 16 grams. So in 2 moles there are 32 grams
