Answer:
975.56×10²³ molecules
Explanation:
Given data:
Number of molecules of C₂H₆ = 4.88×10²⁵
Number of molecules of CO₂ produced = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Number of moles of C₂H₆:
1 mole = 6.022×10²³ molecules
4.88×10²⁵ molecules×1mol/6.022×10²³ molecules
0.81×10² mol
81 mol
Now we will compare the moles of C₂H₆ with CO₂.
C₂H₆ : CO₂
2 : 4
81 : 4/2×81 = 162 mol
Number of molecules of CO₂:
1 mole = 6.022×10²³ molecules
162 mol ×6.022×10²³ molecules / 1 mol
975.56×10²³ molecules
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
1) The forward reaction is N2 (g) + O2 (g) → 2NO
(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.
2) The equiblibrium equation is
N2 (g) + O2 (g) ⇄ 2NO
3) Then, the reverse reaction is
2NO → N2(g) + O2(g)
Answer: 2NO → N2(g) + O2(g)
On oxidation of aldehydes produces carboxylic acid functional group.
The product of oxidation of aldehydes depends upon whether the reaction occur in acidic medium or alkaline condition.
If oxidation of aldehydes occurs under acidic condition the product is carboxylic acid but if oxidation of aldehydes occurs under alkaline condition then reduce as well as oxidized product obtained which is known as disproportional product.
The oxidation of aldehydes occur through potassium dichromate, potassium permanganate or many more. The oxidation of aldehydes in the presence of base is known as cannizzaro's reaction.
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