Explanation:
F = k.m1.m2/r²
m1 and m2: masses of objects
r: distance between the objects
k: a constant
Explanation:
As the charge of all electrons are equal, the repulsive force exerted by each of them is also going to be equal. So, as K has more electrons repulsing its valence electron than Na, it has greater electron shielding.
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Explanation:
In this reaction, the reactants are Li and N2. The product is Li3N
So we have;
Li + N2 → Li3N
Upon balancing, we have;
6Li + N2 → 2 Li3N
The sum of the coefficients is 6 + 1 + 2 = 9
