Answer: Hello, There! Your Answer is Below
2.4 x 1024 ions
Explanation:
220 g of CaCl₂ = X moles
Solving for X,
X = (220 g × 1 mol) ÷ 110.98 g
X = 1.98 moles
As,
1 mole contained = 1.20 × 10²⁴ Cl⁻ Ions
Then,
1.98 mole will contain = X Cl⁻ Ions
Solving for X,
X = (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol
X = 2.38 × 10²⁴ Cl⁻ Ions
Hope this Helps!
Have a great DAy!
`August~
The molecular mass of sucrose is 342.3<span> grams per mole (g/mol).</span>
<span>divide the 201g by the mol mass of the compound. Just add up the masses of the various element</span>
1 L ------- 1000 cm³
1.45 L ----- ???
1.45 * 1000 = 1450 cm³ ( volume )
Density = 0.710 g/cm³
mass = in Kg
m = D * V
m = 0.710 * 1450
m = 1029.5 g
1 Kg ------- 1000 g
kg -------- 1029.5 g
mass = 1029.5 / 1000
mass = 1.0295 Kg
hope this helps!
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.
