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xxMikexx [17]
3 years ago
13

What is the vapor pressure of an aqueous solution made up of 60.0g of urea (CH4N2O) in 180g of water? the vapor pressure of wate

r at this temperature is 0.038atm.
please answer truthfully and explain how you got this answer because i dont know and my head hurts.
also the choices are
0.020atm
0.0025atm
0.035atm
0.00055atm
Chemistry
1 answer:
djyliett [7]3 years ago
7 0
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
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2 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0
3 years ago
3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.
igor_vitrenko [27]
Volume percent = Volume of solute
                              ----------------------------------
                                Volume of the solution
  
                  2                    Volume of the solute
               -------   =           ------------------------------
                100                               250
                     
         Volume of the solute = 2 x 250
                                                ------------
                                                  100         

                                             =   5 mL.

Hope this helps!




                            




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Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope
morpeh [17]

<u>Answer:</u>

<u>For a:</u> The isotopic representation of iodine is _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic representation of cesium is _{55}^{137}\textrm{Cs}

<u>For c:</u> The isotopic representation of strontium is _{38}^{52}\textrm{Sr}

<u>Explanation:</u>

The isotopic representation of an atom is: _Z^A\textrm{X}

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

  • <u>For a:</u>

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is _{53}^{131}\textrm{I}

  • <u>For b:</u>

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is _{55}^{137}\textrm{Cs}

  • <u>For c:</u>

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is _{38}^{52}\textrm{Sr}

3 0
3 years ago
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