The reaction is of order three with respect to the reactant.
<h3>Explanation</h3>
The rate of a reaction of order n about a certain reactant is proportion to the concentration of that reactant raised to the n-th power. This is true only if concentrations of any other reactants stay constant in the whole process.
In other words, Rate = constant × [Reactant]ⁿ, Rate ∝ [Reactant]ⁿ. (The symbol "∝" reads "proportional to".)
In this question,
[4 × Reactant]ⁿ ÷ [Reactant]ⁿ = 64.
In other words, 4ⁿ = 64, where n is the order of the reaction with respect to this reactant.
It might take some guesswork to find the value of n. Alternatively, n can be solved directly with a calculator using logarithms. Taking natural log of both sides:
.
Evaluating 
on Google or on a calculator with support for ln (the natural log) will give the value of n- no guesswork required.
n = 3. Therefore, the reaction is of order three with respect to this reactant.
24g 4Al+3O2=2Al2O3
27g x(g)
108 96
X=24g
Answer:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
Explanation:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
This is a single displacement reaction.
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
Answer
I think there should be two poles, a north pole and a south pole. so this statement is wrong
This is a freezing point depression problem, so it will use the equation:
ΔT = i Kf<span> m
</span>
i = 1 (naphthalene does not dissociate further when dissolved)
Kf = 7.10 C/m (a constant for paradichlorobenzene, which you'd be given)
m = moles of solute / kg of solvent; moles of solute = 4.78 / 128.2 = 0.0373; kg of solvent = 0.032; m = 0.0373 / 0.032 = <span>1.166m
</span>ΔT = 1 x 7.10 x 1.166 = 8.279 C
This means the normal freezing point of pure paradichlorobenzene is decreased by 8.279 C; the normal freezing point (again, something you'd be given) is 53.5 C, so the new freezing point would be 53.5 - 8.279 = 45.221 C.
