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katrin2010 [14]
2 years ago
15

What is the solubility of La(IO₃)₃ in a solution that contains 0.100 M IO₃⁻ ions? (Ksp of La(IO₃)₃ is 7.5 × 10⁻¹²)

Chemistry
1 answer:
love history [14]2 years ago
3 0

<u>Solution and Explanation:</u>

[La3+] = 0.1 M

<u>At the equilibrium: </u>

La(IO3)3 <---->     La3+     +         3 IO3-  

                    0.1 +s             3s      

\mathrm{Ksp}=[\mathrm{La} 3+][\mathrm{IO} 3-]^{\wedge} 3

7.5 * 10^{\wedge}-12=(0.1+s) *(3 s)^{\wedge} 3

Since Ksp is small, s can be ignored as compared to the 0.1

The above shown expression thus becomes:

7.5 * 10 \wedge-12=(0.1) *(3 s) \wedge 3

7.5 * 10^{\wedge}-12=0.1 * 27(\mathrm{s})^{\wedge} 3

s=1.406 * 10^{\wedge}-4 \mathrm{M}

<u>Answer: 1.4*10^-4 M </u>

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marin [14]

Answer:

Because  two minerals can share property's so you need more then one to tell them apart. Hope this helped! :)

Explanation:

8 0
3 years ago
Predict the products of La(s) + O2(aq) -&gt;
mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

<em>I hope this </em><em>helps </em><em>you</em><em> </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Enjoy </em><em>your</em><em> </em><em>studies</em>

3 0
2 years ago
A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 0
3 years ago
Nitrous oxide (n2o), or laughing gas, is commonly used as an anesthetic in dentistry and surgery. how many moles are present in
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Answer is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.

V(N₂O) = 8.4 L.

V(N₂O) = n(N₂O) · Vm.

Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.

n(N₂O) = 8.4 L ÷ 22.4 L/mol.

n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>

5 0
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aliya0001 [1]
I have attached the answer. hopefully, i read the problems correctly. let me know if I did not.

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