N₂ + 3H₂ ⇒ 2NH₃
doesnt matterN₂ + 6.64H₂ ⇒ 2NH₃
(6.64H₂/3H₂) x (2NH₃) =4.4266667
rounded to sig figs= 4.43
The equation is
W = C/F
W= 3.00 x 10^8 m/sec
——————————
6.165 x 10^14 Hz
W= 4.87 x 10^-7 m
Energy is
E=hF
E= (6.626 x 10^-34 Jxsec )(6.165 x 10^14 Hz)
E= 4.085 x 10^-19 J
The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
1) List the reactants: sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇).
Reactants undergo change during a chemical reaction.
2) List the products: water (H₂O), carbon dioxide (CO₂) and sodium citrate (Na₃C₆H₅O₇).
Products are the substances formed from chemical reactions.
3) The balanced chemical equation:
3NaHCO₃ + H₃C₆H₅O₇ → 3H₂O + 3CO₂ + Na₃C₆H₅O₇.
