The grams of aluminium extracted from 5000g of alumina is 2647 grams
<h3>Chemical formula of alumina:</h3>
Let's calculate the molecular mass of Al₂O₃
Al₂O₃ = 27 × 2 + 16 × 3 = 54 + 48 = 102 g/mol
Therefore,
102 g of Al₂O₃ = 54 g of aluminium
5000g of Al₂O₃ = ?
mass of aluminium produced = 5000 × 54 / 102
mass of aluminium produced = 270000 / 102
mass of aluminium produced = 2647.05882353
mass of aluminium produced = 2647 grams
learn more on mass here: brainly.com/question/14627327
The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
All are CORRECT except (d)
Answer: The charge on the plates are 88.4 picafarad
Explanation:The equation used in measuring charge in a plate is given as:
C=Q/V =E A/D
Where E= dielectric content
A= Area of plates
d= distance between plates
Using dielectric constant for Air=8.84×10-12F/m
A=100cm2=0.01m2
d=10mm=0.001m
C= 8.84×10-12×0.01/0.001
C= 88.4 picafarad
No. of atoms=mols*avagadros no.
N=n*No
N=17 * 6.022 *10^23
No. Of atoms=(17) (6.022*10^23)
