<u>Answer:</u> The edge length of the unit cell is 0.461 nm
<u>Explanation:</u>
We are given:
Atomic radius of iridium = 0.163 nm
To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:
Putting values in above equation, we get:
Hence, the edge length of the unit cell is 0.461 nm
Answer:
elements in same group have same valence electron.
period of element is equal to valence shell
Explanation:
if the elements are in a same group then they will be having same number of valence electron.
the period of an element in periodic table is equal to the valence shell of the element. that is if the valence electron are in 3 rd shell then the element will be in third period.
Answer:
2.01 moles of P → 1.21×10²⁴ atoms
2.01 moles of N → 1.21×10²⁴ atoms
4.02 moles of Br → 2.42×10²⁴ atoms
Explanation:
We begin from this relation:
1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br
Then 2.01 moles of PNBr₂ will have:
2.01 moles of P
2.01 moles of N
4.02 moles of Br
To determine the number of atoms, we use the relation:
1 mol has NA (6.02×10²³) atoms
Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms
2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms
4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms
The chemical equation given is:
<span>2x(g) ⇄ y(g)+z(s)</span>
Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of products y(g) and z(s) will be obtained at equilibrium.
Justification:
As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.
The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
The correct option is B.
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