I think that it is half are required to react with 40 g of HF.
Answer:
0.294 M
Explanation:
The computation of the final molarity of acetate anion is shown below:-
Lead acetate = Pb(OAc)2
Lead acetate involves two acetate ion.
14.3 gm lead acetate = Mass ÷ Molar mass
= 14.3 g ÷ 325.29 g/mol
= 0.044 mole
Volume of solution = 300 ml.
then
Molarity of lead is
= 0.044 × 1,000 ÷ 300
= 0.147 M
Therefore the molarity of acetate anion is
= 2 × 0.147
= 0.294 M
Answer:
[HI] = 0.097 M
Explanation:
Let's consider the following reaction.
2 HI(g) ⇄ H₂(g) + I₂(g)
The order of reaction for HI is 1. Thus, we can calculate the concentration of HI ([HI]) at certain time using the following expression:
ln [HI] = ln [HI]₀ - k. t
where,
[HI]₀ is the initial concentration of HI
k is the rate constant
t is the time elapsed
When [HI]₀ = 0.440 M and t = 0.210 s, the concentration of HI is
ln [HI] = ln (0.440) - 7.21 s⁻¹ × 0.210 s
ln [HI] = -2.33
[HI] = 0.097 M
Answer:
B. N₂O₄.
Explanation:
- We will suppose the fuel has the formula NxOy.
- <em>The molar mass of the fuel = (x . atomic mass of N) + (y . atomic mass of O) = 92.0 g/mole.</em>
<em>The % composition of N = 30.43 %</em>
- The % composition of N = (x.atomic mass of N)/(molar mass of fuel) x 100.
- then x = (% composition of N)(molar mass of fuel) / 100(atomic mass of N) = (30.43 %)(92.0 g/mole) / 100(14.00 g/mole) = 1.9996 ≅ 2.0.
<em><u>By the same way; </u></em>
<em>The % composition of O = 69.57 %</em>
- The % composition of O = (y.atomic mass of O)/(molar mass of fuel) x 100
- then y = (% composition of O)(molar mass of fuel) / 100(atomic mass of O) = (69.57 %)(92.0 g/mole) / 100(16.00 g/mole) = 4.0.
<u><em>So, the formula of the compound is N₂O₄.</em></u>
Binary compounds consist of only two distinct elements, regardless of whether the compound is ionic or molecular. Water is a binary compound, as are calcium chloride, ammonia, and potassium iodide.
<span>An ionic binary compound consists of cations of one element and anions of another. KI is an ionic binary compound, composed of K cations and I anions. </span>
<span>A molecular binary compound does not consist of discrete ions, but of molecules. H2O is molecular, as is NH3.</span>