Answer:
Kc for this reaction is 0.43
Explanation:
This is the equilibrium:
N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)
And we have all the concentration at equilibrium:
N₂: 0.25M
H₂ : 1.3M
NO: 0.33M
H₂: 1.2M
They are ok, because they are in MOLARITY. (mol/L)
Let's make the expression for Kc
Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)
Kc = (0.33² . 1.2²) / (0.25 . 1.2²)
Kc = 0.4356
In two significant digits. 0.43
Etaphase is a stage during the process of cell division (mitosis or meiosis). Usually, individual chromosomes cannot be observed in the cell nucleus. However, during metaphase of mitosis or meiosis the chromosomes condense and become distinguishable as they align in the center of the dividing cell.
Answer:The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol).
The molar mass of a compound can be calculated by adding the standard atomic masses (in g/mol) of the constituent atoms.
Explanation:
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)
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