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
Answer:
CaO + SO 3 → CaSO. 4
This is an acid-base reaction (neutralization): CaO is a base, SO 3 is an acid.
Answer:
An educated guess based on what you already know.
Explanation:
You know 1 tbsp is 15 milliliters so you would multiply 15 and 3 together. 15 x 3 =45
So there are 45 milliliters in 3 tbsp.
Answer:
3.8 x 10⁵
Explanation:
For the equilibrium : 3NO(g) ⇌ N2O(g) + NO2(g), the equilibrium constant in the terms of the concentrations of the gases in mol/L is
Kc = (NO) (N2O)/ (NO) ³ where (NO), (N2O) , (NO2) are the concentrations of the gases in mol/L . So
K= (x mol/ 1 L) (x mol/1L) / (7.5 x 10⁻⁶ mol /1 L) ³
x = mol of NO and NO2 at equilibrium
we have that
K = x²/ 7.5 x 10⁻⁶ = 1.9 x 10¹⁶
x = √ (7.5 x 10⁻⁶ x 1.9 x 10¹⁶) = 3.8 x 10⁵
∴ (N2O) = 3.8 x 10⁵
