Divide 50 miles an hour or mph by 0.621, should give you 80.51 kilometers an hour or km/hr, this being the speed limit on that street in km/hr. If you're going 120 km/hr on an 80.51 kilometers an hour road you are speeding, as you're above the speed limit.
Answer:
grams of oxygen = 21.40 g
Explanation:
octane → C8H18
The chemical reaction needs to be represented with a chemical equation before solving. The equation also need to be balanced.
C8H18 + O2 → CO2 + H2O
Balanced equation
2C8H18 + 25O2 → 16CO2 + 18H2O
Molar mass of octane = 12 × 8 + 18 = 96 + 18 = 114 g
Mass of octane in the chemical equation = 2 (114) = 228 g
molar mass of oxygen = 32 g
mass of oxygen in the chemical equation = 25 × 32 = 800 g
if 228 g of octane react with 800 g of oxygen
6.1 g of octane will require ? grams of oxygen
cross multiply
grams of oxygen = 6.1 × 800/228
grams of oxygen = 4880
/228
grams of oxygen = 21.4035087719
grams of oxygen = 21.40 g
Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] ·
[H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.
Answer:
Explanation:
7 carbon atoms, 6 hydrogen atoms,9 NO2 atoms
7+6+9=22
2(22)=44
44 atoms