(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
Answer:
pH 9,8 is likely to work best for this separation
Explanation:
Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).
For lysine, PI is:
9,8
For arginine:
10,75
At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.
Thus, <em>pH 9,8 is likely to work best for this separation</em>
<em></em>
I hope it helps!
Answer:
17.57kg of
and its percentage yield is 81.0%
Explanation:
Through the reaction you can get the theoretical amount of
that must be produced.

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage
=81.0%
Answer:

Explanation:
Mol of NaI = 0.405 mol
Molarity of solution = 0.724 M
Molarity is given by

The required volume is
.
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M