2.) Average atomic mass =Σ (abundance x molar mass) /100
= (12.64 x 302.04 + 18.23 x 304.12 + 69.13 x 305.03) /100
=304.486 u(Dalton) . This is avg atomic mass.
Answer:
Na the valency electrons is 1
The ion charge is positive 1
K the valency electrons is 1
The ion charge is positive 1
Explanation:
Mg the valency electron is 2 The ion charge is positive 2
Ca the valency electron is 2 The ion charge is positive 2
Al the valency electron is 3 The ion charge is positive 3
O the valency electron is 2 The ion charge
is negative 2
F the valency electron is 1 The ion charge is negative 1
Cl the valency electron is 1 The ion charge
is negative 1
N the valency electron is 3 The ion charge is negative 3
S the valency electron is 2 The ion charge is
negative 2
Answer:
32.23 to 4 significant figures.
Explanation:
The molar mass of the element is the mass of 6.022 * 10^23 atoms (Avogadro's number).
So by proportion it is 6.022 * 10^23 * 3.88 / 7.25 * 10^22
= 32.23 to 4 significant figures.
Explanation:
1. Electrons surround the nucleus in defined regions called orbits.
2. The shells further away from the nucleus are larger and can hold more electrons.
3. The shells closer to the nucleus are smaller and can hold less electrons.
4. The closest shell (closest to the nucleus) can hold a maximum of two electrons.
5. Once the first shell is full, the second shell begins to fill. It can hold a maximum of eight electrons.
6. Once the second shell is full, the third shell begins to fill.
7. Once the third shell contains Eighteen electrons, the fourth shell begins to fill.
8. The arrangement of electrons in shells around the nucleus is referred to as an atom's electronic configuration.
According to <span>Gay-Lussac's Law the temperature and Pressure are directly proportional to each other if the amount and volume of given gas are kept constant.
Mathematically for initial and final states it is expressed as,
P</span>₁ / T₁ = P₂ / T₂ ----- (1)
Data Given;
P₁ = 1.5 atm
T₁ = 35 °C + 273 = 308 K
P₂ = ?
T₂ = 0 °C + 273 = 273 K
Solving Eq. 1 for P₂,
P₂ = P₁ T₂ / T₁
Putting values,
P₂ = (1.5 atm × 273 K) ÷ 308 K
P₂ = 1.32 atm
Result:
As the temperature is decreased so the pressure also decreases from 1.5 atm to 1.32 atm. Therefore the bag will contract.
