When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
<span>Molarity is expressed as
the number of moles of solute per volume of the solution. We calculate as follows:
2.80 g ( 1 mol / 56.11 g ) = 0.05 mol KOH
Molarity = 0.05 mol KOH / 750 mL ( 1 L / 1000 mL )
Molarity = 0.07 M
Hope this answers the question. Have a nice day.</span>
Answer:
C.
Explanation:
The electronic configuration of N (7 electrons): 1s² 2s² 2p³.
The orbital 1s is filled with two electrons and their spinning direction is opposite and also electrons of 2s.
3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.
So, the right choice is c.
A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.
B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.
D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.
So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!
