There are 3 moles of
<span>per 1 mole of salt and 1 mole of
</span>per mole of salt, the total ionic concentrations must be
of
, and
of
Answer:
1.7 * 10^-5
Explanation:
1- get the number of moles of PbCl2:
number of moles = mass / molar mass
number of moles = 0.45 / 278.1 = 1.618 * 10^-3 moles
2- get the concentration of Pb2+:
molarity = number of moles of solute / volume of solution in liters
molarity = (1.618 * 10^-3) / (0.1) = 0.0162 M
3- getting concentration of Cl-:
<span>PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)
</span>We can note that:
For a certain amount of Pb2+ formed, twice this amount of Cl- is formed.
This means that:
for 0.0162 M of Pb2+, 2*0.0168 = 0.0324 M of Cl- is formed
4- getting Ksp:
Ksp = [Pb2+][Cl-]²
Ksp = (0.0162)*(0.0324)²
Ksp = 1.7 * 10^-5
Hope this helps :)
Answer:
4.13X10^3= 4130 in the expanded form
Answer:
The elements can be classified as metals, nonmetals, or metalloids. Metals are good conductors of heat and electricity, and are malleable (they can be ... and electricity, and are not malleable or ductile; many of the elemental nonmetals are ... under certain circumstances, several of them can be made to conduct electricity.
Hope this helps!
Explanation:
Answer:
The endpoint volume is 50.52 ± 0.14 mL
Explanation:
In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:
V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)
V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL
V = 50.52 ± 0.14 mL
It is necessary to consider the sum of the errors too.