Answer:
to make sure that the source is reliable.
Explanation:
because I am smart
Answer:
26.8g
Explanation:
The formula of the compound given is:
Cu₂CrO₄
Given:
Number of moles = 0.11
To find the mass, we use the expression below:
Mass = number of moles x molar mass
Molar mass of Cu₂CrO₄ = 2(63.6) + 52 + 4(16) = 243.2g/mol
Now insert the parameters and solve;
Mass = 0.11 x 243.2 = 26.8g
a) Alkali metals
=> group 1
=> Li: 1s2 2s => 1s
Na: [Ne] 3s => 3s
K: [Ar] 4s => 4s
Rb: [Kr] 5s => 5s
Cs: [Xe] 6s => 6s
Fr: [Rn] 7s => 7s
=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
b) Alkaline earth metals
=> group 2 => you have to add 1 electron to the alkaly metal of the same row.
=> Be: [He] 2s2 => 2s2
Mg: [Ne] 3s2 => 3s2
Ca: [Ar] 4s2 => 4s2
Sr: [Kr] 5s2 => 5s2
Ba: [Xe] 6s2 => 6s2
Ra: [Rn[ 7s2 => 7s2
=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7
c) halogens
=> group 7
=> F: [He] 2s2 2p5 => 2s2 2p5
Cl: [Ne] 3s2 3p5 => 3s2 3p5
Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
At: [Xe] 4f14 5d10 6s2 6p5
=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7
d) Noble gases
=> group 8
I will show only the outer shell which is what is requested
=> He: 1s2
Ne: ... 2s2 2p6
Ar: ... 3s2 3p6
Kr: ... 4s2 4p6
Xe: ... 5s2 5p6
Rn: ... 6s2 6p6
=> the outer electron configuration is ns2 np6, except for He for which it is 1s2
A - its condensation and gas particles have a higher kinetic energy
