<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
We can solve this without a concrete formula through dimensional analysis. This works by manipulating the units such that you end up with the unit of the final answer. Manipulate them by cancelling units that appear both in the numerator and denominator side. As a result, we must be left with the units of g. The current in A or amperes is equivalent to amount of Coulombs per second. Since this involves Coulombs, we will use the Faraday's constant which is 96,500 C/mol electron. The reaction is:
Cr³⁺(aq) + 3e⁻ --> Cr(s)
This means that for every 3 moles of electron transferred, 1 mole of Chromium metal is plated. The molar mass of Cr: 52 g/mol. The solution is as follows:
Mass of Chromium metal = (8 C/s)(60 s/1 min)(160 min)(1 mol e⁻/96,500 C)(1 mol Cr/3 mol e)(52 g/mol)
<em>Mass of Chromium metal = 13.79 g</em>
Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
The Lewis structure of
Diimide (N₂H₂) is shown below.
In this molecule two Nitrogen atoms attached to each other through a
double bond are further attached to one one Hydrogen atom. Also, each Nitrogen atom carries one
non-binding electron pair (
Lone Pair) (Highlighted RED).
Result: Option-C (<span>each nitrogen has one nonbinding electron pair) is the correct answer.</span>
