The mass of ethanol (alcohol) = 538.016 g
<h3>Further explanation</h3>
Given
Reaction
C₁₂H₂₂O₁₁(aq)+H₂O(I) = CH₃CH₂OH(aq)+CO₂(g)
1 kg of sugar = 1000 g
Required
The mass of ethanol
Solution
Balanced equation
C₁₂H₂₂O₁₁(aq)+H₂O(I) = 4CH₃CH₂OH(aq)+4CO₂(g)
mol of sugar (MW = 12.12+22.1+11.16=342 g/mol) :
mol = mass : MW
mol = 1000 g : 342 g/mol
mol = 2.924
From the equation, mol ratio of C₁₂H₂₂O₁₁ : CH₃CH₂OH = 1 : 4, so mol ethanol =
= 4 x mol sugar
= 4 x 2.924
= 11.696
Mass of ethanol(MW=46 g/mol) :
mass = mol x MW
mass = 11.696 x 46
mass = 538.016 g
Answer:
I kinda forgot. I'm sorry if I didn't answer your question.
Explanation:
Answer:
drip drip water splash (gas)
Explanation:
Thank you for the free 15 points .
Answer:
Explanation:
mass of one virus = 9.0 x 10⁻¹² mg
mass of one mole = 6.02 x 10²³ x mass of one virus
= 6.02 x 10²³ x 9.0 x 10⁻¹²
= 54.18 x 10¹¹ mg
= 54 x 10⁸ g .
= 54 x 10⁵ kg .
b )
let n be no of moles of virus that will be equal to weight of oil tanker
n x 54 x 10⁵ = 3 x 10⁷
n = 5.5555
rounding off to 2 significant figure
5.6 moles Ans .
