Answer:
Explain more please
Explanation:
you have to at the statements
Answer:
Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)
Explanation:
<em>Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO₃ and HI are mixed.</em>
When MgSO₃ reacts with HI they experience a double displacement reaction, in which the cations and anions of each compound are exchanged, forming H₂SO₃ and MgI₂. At the same time, H₂SO₃ tends to decompose to H₂O and SO₂. The complete molecular equation is:
MgSO₃(aq) + 2 HI(aq) ⇄ MgI₂(aq) + H₂O(l) + SO₂(g)
In the complete ionic equation, species with ionic bonds dissociate into ions.
Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)
Answer:
The value of y = 5.1478
Explanation:
The linear equation is an equation obtained when a linear polynomial is equated to zero. When the solution obtained on solving the equation is substituted in the equation in place of the unknown, the equation gets satisfied.
The given equation: 5.3 x 10- (y)(2y) = 0
⇒ 53 - 2y² = 0
⇒ 2y² = 53
⇒ y² = 53 ÷ 2 = 26.5
⇒ y = √26.5 = 5.1478
Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of 
= 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of = 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
