Explanation:
The nomenclature of ionic compounds is given by:
1. Positive is written first. And if the metal atom has various oxidation states then its oxidation state is to be mentioned in brackets with help of roman numbers
2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
1.) : Calcium chloride (Correct)
In the given compound, calcium has oxidation state of +2 and chlorine has oxidation state of -1.The name is correct.
2.) : Copper (II) oxide (Incorrect)
In the given compound,copper has the oxidation state of +1 and oxygen has oxidation state of -2.So ,the correct name will be Copper (I) oxide.
3.) : Lead (II) sulfide (Incorrect)
In the given compound, lead has the oxidation state of +4 and sulfur has oxidation state of -2. So ,the correct name will be Lead (IV) sulfide.
Answer:
Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + 2NaNO₃
The coefficients are: 1, 1, 1, 2
Explanation:
Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + NaNO₃
The above equation can be balance as follow:
Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + NaNO₃
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 2 before NaNO₃ as shown below:
Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + 2NaNO₃
Now the equation is balanced.
The coefficients are: 1, 1, 1, 2
Answer:
a) [ Ca2+ ] = 3.347 E-4 mol/L
b) [ Ca2+ ] = 1.5 E-8 mol/L
Explanation:
S S 2S......in the equilibrium
⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S )² = 4S³
⇒ S = ∛ ( 1.5 E-10 / 4 )
⇒ S = ∛ 3.75 E-11
⇒ S = 3.347 E-4 mol/L
⇒ [ Ca2+ ] = S = 3.347 E-4 mol/L
b) NaF ↔ Na+ + F-
0.10 M 0.10 0.10
S S 2S + 0.10
⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S + 0.10 )²
∴∴ the Concentration: 0.10 M >>>> Ksp ( 1.5 E-10 ), son we can despise S as adding.
⇒ 1.5 E-10 = S * ( 0.10 )² = 0.01 S
⇒ S = 1.5 E-10 / 0-01
⇒ S = 1.5 E-8 mol/L
⇒ [ Ca2+ ] = S = 1.5 E-8 mol/L
Answer:red green and blue hope it helps!,!!!!!!!
Explanation:
