Answer:2 AL + FEN2 = 2 ALN + FE
Explanation: AL might be an improperly capitalized: Al. One of your compounds is AL (A and L). Did you mean Al (aluminum)?
FEN2 might be an improperly capitalized: FeN2. ALN might be an improperly capitalized: AlN
FE might be an improperly capitalized: Fe
From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
This reaction is known as
Acidic Cleavage of Ether.
Ether when treated with strong acid like
HI, the oxygen atom of ether picks the proton making a positive charge on oxygen. The
Iodide conjugate base formed acts as a
nucleophile and attacks on the carbon next to positive oxygen and forms
Alkyl Halide. This is a type of
SN² reaction. The complete reaction with missing atoms, charges and arrows is shown below,
Answer:
0.0010m SO₄²⁻
Explanation:
The freezing point depression due the addition of a solute into a pure solvent follows the equation:
ΔT = Kf×m×i (1)
<em>Where ΔT are °C that freezing point decreases (273.15K - 272.47K = 0.68K = 0.68°C). Kf is the constant of freezing point depression (1.86°C/m), m is molality of the solution (0.1778m) and i is Van't Hoff factor.</em>
Van't Hoff factor could be understood as in how many one mole of the solute (sulfuric acid, H₂SO₄), is dissociated.
H₂SO₄ dissociates as follows:
H₂SO₄ → HSO₄⁻ + H⁺
HSO₄⁻ ⇄ SO₄²⁻ + H⁺
<em>Not all HSO₄⁻ dissociates.</em>
1 Mole of H₂SO₄ dissociates in 1 mole of H⁺+ 1 mole of HSO₄⁻ + X moles of SO₄²⁻= 2 + X
Replacing in (1):
0.68°C = 1.86°C/m×0.1778m×i
2.056 = i
Moles of SO₄²⁻ are 2.056 - 2 = 0.056moles SO₄²⁻.
If 1 mole has a concentration of 0.1778m, 0.056moles are:
0.056moles ₓ (0.1778m / 1mole) =
<h3>0.0010m SO₄²⁻</h3>
Answer:
1. conduction
2. radiation
3.convection
4. conduction?
5. radiation
6. convection
7. convection
8. convection
9. radiation
10.radiation
(sorry if they aren't all correct)
