Mass number is the sum of protons and neutrons.
Atomic number is the total number of protons present.
Mass number = protons + neutrons
Atomic number = protons present.
From the above equations we can conclude that,
Atomic number= Mass number- neutrons.
Thus we can find out the atomic number by subtracting the number of neutrons from the mass number.
Answer:
[He]: 2s² 2p⁵.
[Ne]: 3s².
[Ar]: 4s² 3d¹⁰ 4p².
[Kr]: 5s² 4d¹⁰ 5p⁵.
[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².
Explanation:
- Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.
He contains 2 electrons fill 1s (1s²).
So, [He] can be written before the electronic configuration of 2s² 2p⁵.
Ne contains 10 electrons fill (1s² 2s² 2p⁶).
So, [Ne] can be written before the electronic configuration of 3s².
Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).
So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².
Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).
So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.
Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).
So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².
Answer:
Final Volume = 5.18 Liters
Explanation:
Initial Condition:
P1 = 789 mm Hg x (1/760) atm /mm Hg = 1.038 atm
T1 = 22° C = 273 + 22 = 295 K
V1 = 4.7 L
Final Condition:
P2 = 755 mm Hg x (1/760) atm /mm Hg = 0.99 atm
T2 = 37° C = 273 + 37 = 310 K
V2 = ?
Since, (P1 x V1) / T1 = (P2 x V2) / T2,
Therefore,
⇒ (1.038)(4.7) / 295 = (0.99)(V2) / 310
⇒ V2 = 5.18 L (Final Volume)
Answer: 1.6L
Explanation:
V1 = 1.50 L,
V2 =?
n1 = 3mol
n2 = 3 + 0.2 = 3.2mol
From PV = nRT
V1 /n1 = V2/n2
1.5/3 = V2 /3.2
V2 = (1.5/3/) x 3.2 = 1.6L
