hydrocarbon is ethene which is used to test for saturation and it undergoes addition reaction
Answer:
a.) 22.4 L Ne.
Explanation:
It is known that every 1.0 mol of any gas occupies 22.4 L.
For the options:
<em>It represents </em><em>1.0 mol of Ne.</em>
<em />
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 20 L.
The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 2.24 L.
<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>
- So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.
Answer:
Mass of Sodium = 574.75 g
Mass of Chlorine = 886.25 g
Explanation:
The balance chemical equation for the synthesis of NaCl is,
2 Na + Cl₂ → 2 NaCl
Step 1: <u>Find out moles of each reactant required,</u>
According to balance chemical equation,
2 moles of NaCl is produced by = 2 moles of Na
So,
25 moles of NaCl will be produced by = X moles of Na
Solving for X,
X = 25 mol × 2 mol / 2 mol
X = 25 moles of Na
Similarly for Cl₂,
According to balance chemical equation,
2 moles of NaCl is produced by = 1 mole of Cl₂
So,
25 moles of NaCl will be produced by = X moles of Cl₂
Solving for X,
X = 25 mol × 1 mol / 2 mol
X = 12.5 moles of Cl₂
Step 2: <u>Convert each moles to mass as;</u>
Mass = Moles × Atomic Mass
For Na,
Mass = 25 mol × 22.99 g/mol
Mass = 574.75 g
For Cl₂,
Mass = 12.5 mol × 70.90 g/mol
Mass = 886.25 g
Stop littering and polluting
