The following are the answers to the different questions:
<span>The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?
A. 101.53° C
Which of the following solutions will have the lowest freezing point?
D. 1.0 mol/kg magnesium fluoride (MgF2)
Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?
A. sodium iodide (NaI)
Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?
B. 96.3 kPa
Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?
D. 108.1°C</span>
We are given –
- Mass of
is 57.1 g and we are asked to find number of moles present in 57.1 g of 






____________________
Now,Let's calculate the number of moles present in 57.1 g of 






__________________________________
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
For an non spontaneous reaction between silver (Ag) and copper (Cu) and their ions, Cu is the oxidizing agent while Ag+ is the reducing agent,
The following reactions will take place;
Anode Cu = Cu+2 + 2e- E= +0.34 volts
Cathode; Ag+ + e = Ag E = +0.80 volts
The net reaction will be Cu + 2Ag+ = Cu+2 + 2Ag
Thus, the voltage will be
= +0.80 - (+0.34)