Answer:
Relation between , molality and temperature is as follows.
T =
It is also known as depression between freezing point where, i is the Van't Hoff factor.
Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.
i for = 3
i for glucose = 1
i for NaCl = 2
Depression in freezing point will have a negative sign. Therefore, d
depression in freezing point for the given species is as follows.
=
=
=
Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.
Glucose < NaCl <
Explanation:
To get the answer you use the Law of Raoult.
Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.
ΔP = Pa * Xa
Here Pa = 0.038 atm
And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b
Na = mass of urea / molar mass of urea = 60 g / (molar mass of CH4N2O)
molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol
Na = 60 g / 60 g/mol = 1 mol
Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol
Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091
ΔP = Pb * Xa = 0.038 atm * 0.09091 = 0.0035 atm
Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.
Answer: 0.035 atm
Answer:
Hybridization: sp
Electron geometry: linear
Molecular geometry: linear
Explanation:
H₃CCCH can also be written as its Lewis structure which is shown in the figure attached. The figure shows that the central carbon atom makes a single bond with CH₃ and a triple bond with CH. This means that the hybridization of the carbon is sp and both the electron and molecular geometry are linear with an 180° bond angle.
2KClO3 --> 2KClO2 + O2
12 6 (moles)
The ratio of KClO3 and O2 is 2:1. This means 2 moles of KClO3 can create 1 mole of O2. So 12 moles of KClO3 will create 6 moles of O2.
