Imagine a chemist is in the lab and trying to make some chemical reactions happen. In one reaction she reacts chemicals in an exothermic reaction and there is an increase in entropy. A second chemical reaction she is trying to run is endothermic and there is a decrease in entropy. Which of the two reactions is more likely to occur and why?
In order to determine the concentration of ammonium ions in
the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium
nitrate, first calculate the amount of ammonium ions for each solution.<span>
<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 =
0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3
= 0.81 mol NH4+
Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99
mol NH4+. To get the concentration, multiply this to the volume of solution
which is assumed to be additive, such that:</span></span></span>
M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L
sol’n
Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero,
for NiO4-, assuming oxidation number of Ni is x
(x ×1) + (-2 × 4) = -1
x = + 7
Therefore, the oxidation number goes from 0 to +7
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)
Answer:
Iron slowly reacts with oxygen and forms rust. In this case, the reactants are iron and oxygen. The product is rust, or iron oxide. The chemical equation looks like this:
Explanation:
iron + oxygen → iron oxide
