Answer:
The answer is D. Hydroelectric
<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
Answer:
28.20 mL of the stock solution.
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 12.1 M
Volume of diluted solution (V2) = 350.0 mL
Molarity of diluted solution (M2) = 0.975 M
Volume of stock solution needed (V1) =..?
The volume of stock solution needed can be obtained by using the dilution formula as shown below:
M1V1 = M2V2
12.1 x V1 = 0.975 x 350
Divide both side by 12.1
V1 = (0.975 x 350)/12.1
V1 = 28.20 mL.
Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.
Covalent example ch4 where carbon is covalently bonded with all 4 hydrogens
<span>In the electron cloud model, the denser areas represent that there is a great probability that a good number of electrons are ganged up or crowded in that area. The electrons affect the density of some parts of the electron cloud when they condense in those locations.</span>
