Answer:
174.8 g/m is the molar mass of the solute
Explanation:
We must apply colligative property of freezing point depression.
ΔT = Kf . m . i
ΔT = T° freezing pure solvent - T° freezing solution (0° - (-2.34°C) = 2.34°C
Kf = Fussion constant for water, 1.86 °C/m
As ascorbic acid is an organic compound, we assume that is non electrolytic, so i = 1
2.34°C = 1.86°C/m . m
2.34°C / 1.86 m/°C = 1.26 m
This value means the moles of vitamin C, in 1000 g of solvent
We weighed the solute in 250 g of solvent, so let's calculate the moles of vitamin C.
1000 g ___ 1.26 moles
In 250 g ___ (250 . 1.26)/1000 = 0.314 moles
This are the moles of 55 g of ascorbic acid, so the molar mass, will be:
grams / mol ⇒ 55 g/0.314 m = 174.8 g/m
Answer: They would both be pushed backward. (READ EXPLANATION)
Explanation:
Since your question has an error in it, I will provide two answers. If they were standing on their feet, they would not move. if they're both standing on skateboards, they would both move backwards because they have no friction on the ground. if both of them are standing on the same skateboard and pushing against each other, they would not move.
Reaction involved in present electrochemical cell,
At Anode: Zn → Zn^2+ + 2e^2-
At cathode: Zn^2+ + 2e^2- → Zn
Net Reaction: Zn + Zn^2+ ('x' m) → Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2
According to Nernst equation for electrochemical cell,
Ecell = -2.303
![\frac{RT}{nT} log \frac{[Zn^2+]R}{[Zn^2+]L}](https://tex.z-dn.net/?f=%20%5Cfrac%7BRT%7D%7BnT%7D%20log%20%20%5Cfrac%7B%5BZn%5E2%2B%5DR%7D%7B%5BZn%5E2%2B%5DL%7D%20%20)
= 0.014
Given: T =
, F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v
∴ 0.014 = - 2.303
∴ log
=
= -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m
Answer:
1/32 of the original sample
Explanation:
We have to use the formula
N/No = (1/2)^t/t1/2
N= amount of radioactive sample left after n number of half lives
No= original amount of radioactive sample present
t= time taken for the amount of radioactive samples to reduce to N
t1/2= half-life of the radioactive sample
We have been told that t= five half lives. This implies that t= 5(t1/2)
N/No = (1/2)^5(t1/2)/t1/2
Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;
N/No= (1/2)^5
N/No = 1/32
Hence the fraction left is 1/32 of the original sample.
