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:
Explanation:
The acid level has changed
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer:
Anything that can be done to increase the frequency of those collisions and/or to give those collisions more energy will increase the rate of dissolving.
Explanation:
depended on the temperature
Anode- oxidization
Cathode-reduction
