Answer:
The answer to your question is 97 ml
Explanation:
Data
volume = ? ml
concentration = 1.01 M LiNO₃
mass = 6.63 g
Process
1.- Calculate the molecular weight of LiNO₃
LiNO₃ = (1 x 6.94) + (1 x 14) + (16 x 3)
= 6.94 + 14 + 48
= 69 g
2.- Calculate the moles of LiNO₃
69 g ------------------ 1 mol
6.63 g --------------- x
x = (6.63 x 1) / 69
x = 0.096 moles
3.- Calculate the volume
Molarity = moles / volume
-Solve for volume
Volume = Molarity x moles
-Substitution
Volume = 1.01 x 0.096
-Result
Volume = 0.097 l or 97 ml
it is not a prescription drug
Answer:
b) [Kr] 5s² 4d¹⁰ 5p⁵
Explanation:
Properties of iodine:
Iodine is present in group seventeen.
It is halogen element.
Its atomic mass is 127 amu.
Its atomic number is 53.
It is present in solid form.
It is crystalline in nature.
It is very corrosive and has pungent odor.
It can not react with oxygen and nitrogen.
Electronic configuration:
I₅₅ = 1s² 2s² 2p⁶ 3s³ 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁵
Abbreviated electronic configuration:
I₅₅ = [Kr] 5s² 4d¹⁰ 5p⁵
Abbreviated electronic configuration is shortest electronic configuration by using the noble gases.
Answer:
Acetic acid Ka = 1.74 × 10⁻⁵
Trichloroacetic acid Ka = 2 × 10⁻¹
Explanation:
Let's consider the acid dissociation of acetic acid.
CH₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq)
The pKa of acetic acid is 4.76. The acid dissociation constant (Ka) is:
pKa = -log Ka
- pKa = log Ka
Ka = anti log (-pKa)
Ka = anti log (-4.76)
Ka = 1.74 × 10⁻⁵
Let's consider the acid dissociation of trichloroacetic acid.
CCl₃COOH(aq) ⇄ CCl₃COO⁻(aq) + H⁺(aq)
The pKa of trichloroacetic acid is 0.7. The acid dissociation constant (Ka) is:
pKa = -log Ka
- pKa = log Ka
Ka = anti log (-pKa)
Ka = anti log (-0.7)
Ka = 2 × 10⁻¹
