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Answer is: <span>the molarity of the diluted solution 0,043 M.
</span>V(NaOH) = 75 mL ÷ 1000 mL/L = 0,075 L.
c(NaOH) = 0,315 M = 0,315 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,075 L · 0,315 mol/L.
n(NaOH) = 0,023625 mol.
V(solution) = 0,475 L + 0,75 L.
c(solution) = 0,023625 mol ÷ 0,550 L.
c(solution) = 0,043 mol/L.
Answer:
167,3 grams of O2 (g)
Explanation:
According to the following balanced equation:
C3H8 + 5 02 ---> 3 CO2 + 4 H20
We calculate the mass of the moles:
5 mol O2 = 16grams/mol x2x 5= 160 grams/mol
1 mol C3H8= 3 x12 grams/mol + 8 x 1 gram/mol= 44 grams/mol
If 44 grams C3H8 react with--------- 160 grams O2
46 gramsC3H8 react with --------X = (46x160)/44 =167, 3 grams O2
Answer:
(204 g)/(101.96 g/mol) = 2.001 mol. 2(2.001) = 4.002 moles of aluminium in the sample
Explanation:
