Answer:
Solid ionic compounds do not conduct electricity because the ions are held firmly in place. The ions cannot move to conduct the electric current. But when an ionic compound melts, the charged ions are free to move. Therefore, molten ionic compounds do conduct electricity.
The correct unabbreviated electron configuration is as below
Vanadium - 1S2 2S2 2P6 3S2 3p6 3d3 4s2
Strontium - 1s2 2S2 2P6 3S2 3P6 3d10 4S2 4P6 4S2
Carbon =1S2 2S2 2P2
<u><em> Explanation</em></u>
vanadium is in atomic number 23 in the periodic table hence its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2
Strontium is in atomic number 38 in periodic table hence its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4s2
Carbon is in atomic number 6 in periodic table therefore its electron configuration is 1s2 2s2 2p2
Answer:
A precipitate will be produced
Explanation:
The Ksp of AgBr is:
AgBr(s) → Ag⁺ + Br⁻
5.0x10⁻¹³ = [Ag⁺] Br⁻]
<em>Where [] are the concentrations in equilibrium of each ion.</em>
<em />
And if Q is:
Q = [Ag⁺] Br⁻]
<em>Where the concentrations are actual concentrations of each ion</em>
<em />
We can say:
IF Q >= Ksp, a precipitate will be produced
IF Q < Ksp, no precipitate will be produced.
the molar concentrations are:
[AgNO₃] = [Ag⁺] = 0.002M * (50mL / 100mL) = 0.001M
<em>Because 50mL is the volume of the AgNO₃ solution and 100mL the volume of the mixture of both solutions.</em>
[NaBr] = [Br⁻] = 0.002M * (50mL / 100mL) = 0.001M
Q = [0.001M] * [0.001M]
Q = 1x10⁻⁶
As Q > Ksp,
<h3>A precipitate will be produced</h3>
