4)<span>1.24⋅<span>1024</span></span><span> formula units
</span>1 mole Na<span> (sodium)</span>
Answer:
it's-B.....................
Answer:
m = 4.7 μg
Explanation:
Given data:
density of acetone = 60.0 μg/L
Volume = 79.0 mL
Mass = ?
Solution:
Formula:
d = m/v
v = 79.0 mL × 1L /1000 mL
v = 0.079 L
Now we will put the values on formula:
d = m/v
60.0 μg/L = m/0.079 L
m = 60.0 μg/L × 0.079 L
m = 4.7 μg
So health risk limit for acetone = 4.7 μg
Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Answer:
Noble Gases
Explanation:
The Noble Gases have a full valence shell of 8 electrons. They are stable in that sense.
