Answer: 6.Explanation:1) Aluminum

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.
2) Manganesium

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.
3) Balance
Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

4) Net equation
Add the two half-equations:

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.
The right side has 2 Al, 3 Mn, and 2*3 positive charges.
So, the equation is balanced.
5) Count the number of electrons involved.
As you see 2 atoms of aluminum lost 6 electrons (3 each).
That is the answer to the question. 6 electrons will be lost.
D. Protons, Atomic number is the number of protons in an atom.
PbSO₄ partially dissociates in water. the balanced equation is;
PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X
Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm
Answer:
Option A = 2.2 L
Explanation:
Given data:
volume of one mole of gas = 22.4 L
Volume of 0.1 mole of gas at same condition = ?
Solution:
It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.
For 0.1 mole of methane.
0.1/1 × 22.4 = 2.24 L
0.1 mole of methane occupy 2.24 L volume.
Answer:
5.758 is the density of the metal ingot in grams per cubic centimeter.
Explanation:
1) Mass of pycnometer = M = 27.60 g
Mass of pycnometer with water ,m= 45.65 g
Density of water at 20 °C = d =
1 kg = 1000 g


Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g
Volume of pycnometer = Volume of water present in it = V


2) Mass of metal , water and pycnometer = 56.83 g
Mass of metal,M' = 9.5 g
Mass of water when metal and water are together ,m''= 56.83 g - M'- M
56.83 g - 9.5 g - 27.60 g = 19.7 g
Volume of water when metal and water are together = v

Density of metal = d'
Volume of metal = v' =
Difference in volume will give volume of metal ingot.
v' = v - V


Since volume cannot be in negative .
Density of the metal =d'
=