Should precipitation of MgF2 occur if a 17.5mg sample of MgCl2.6H2O is added to 325mL of 0.045M KF? (Ksp MgF2 = 6.4 X 10-9)

Should precipitation of MgF2 occur if a 17.5mg sample of MgCl2.6H2O is added to 325mL of 0.045M KF? (Ksp MgF2 = 6.4 X 10-9)

Solution

We can start by determining [Mg2+], and then the value of Qsp in order to compare it to the value of

Ksp .

Given;

Ksp MgF2 = 6.4 X 10-9

mass of MgCl2.6H2O = 17.5 = 0.0175g

volume = 325mL

[KF] = 0.045M

Calculating the concentration,[Mg2+] ;

mass of MgCl2.6H2O = 17.5 = 0.0175g

Molar mass of MgCl2.6H2O = 203.3g/mol

moles of MgCl2.6H2O = 0.0175/203.3 = 0.0000861 mol

concentration in the given volume = 0.0000861 x (1000/ 325) = 2.645 x 10-4 M

[Mg2+] = 2.645 x 10-4 M

Having the reaction,

MgF2 (s) ------------> Mg+2 (aq) + 2F- (aq)

Qsp = [Mg2+][F-]2 = (2.645 x 10-4)(0.045)2

= 5.356 x 10-7

but Ksp = 6.4 X 10-9

Therefore Qsp > Ksp ;thus, precipitation of MgF2(s) should occur from this solution.

 

In a carbon (graphite)/oxygen fuel cell the cathodic reaction is O2(g) + 4 e- = 2 O2- and the anodic one is CO2(g) + 4 e- = C(s) + 2 O2-. If one uses data from Chapter 10: Data Section, the standard electrode potential difference of the fuel cell at 1000 K and 1 bar is ?

 

Solution

Having the two half cell equations

O2(g) + 4 e- --------->2O2-

CO2(g) + 4 e- ------->C(s) + 2O2- (reverse one equation then add)

------------------------------------------------------------------------------

overall equation C + O2 -------------------> CO2

O2 (g) + 4e + 4 H+ <-------------->H2O----------------------- E0/V =1.229

CO2 (g) + 4e- + 4H+ <------------>C(S) + 2H2O ------------- E0/V = -0.203

C(S) + 2H2O<----------------------->CO2 (g) + 4e- + 4H+ ----- E0/V = +0.203

Now we have

E0 cell = E0 cathode - E0 anode

= E0 reduction - E0 oxidation

= 1.229 - 0.203

= 1.026 V

Thus the standard electrode potential difference of the fuel cell is 1.026V

148.0 grams of ether are mixed with 184.0 grams of ethanol at 80 degrees Celsius. The vapor pressure of ether at 80 degrees Celsius is 4.0 and for ethanol at 80 degrees Celsius is 1.1. Determine the vapor pressure of the solution

Solution

To solve this question,we first need to think about the Raoult's Law which states that the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

This means that we need to find the partial pressure for each component of the mixture the add them .

P ether = X ether x Po ether

P ethanol = X ethanol x Po ethanol

Total vapor pressure = P ether + P ethanol

Where P is the partial vapor pressure of the component;X is the mole fraction of the components and Po is the value of vapor pressures of the components when on their own.

Now calculating the moles

moles of ether = 148/74.12 = 1.97 moles

moles of ethanol= 184/46.07 = 3.99 moles

total moles = 1.97 + 3.99 = 5.96 moles

X ether= moles of ether/total moles

X ether = 1.97 / 5.96 = 0.331

X ethanol = 3.99/5.96 = 0.669

Also given;

Poether = 4.0

Poethanol = 1.1

P ether = 0.331 x 4.0 = 1.324

P ethanol = 0.669 x 1.1 = 0.7359

Total vapor pressure = 1.324 + 0.7359 = 2.0599 = 2.06

Thus the vapor pressure of the solution is 2.06

A wave tank is 120 ft long, 3 ft wide, and 4 ft deep and is fi lled with fresh water to a depth of 3 ft. The wave maker generates a wave which has a wave height of 0.75 ft and wave period of 1.1 s. The density of water is 1.94 slugs/ft3. Evaluate the wave celerity, length, group celerity, energy in one wavelength (EL), and power..

Solution

Wave height,Hm = 0.75ft

water depth = 3ft

For a case of deep water wave

Wave length,λ = g/2π*T2 where T is the period of the wave

= 32.174 /2π *( 1.1)2

=6.195 ft

λ is not much larger than the depth

Thus,

wave celerity ,Cp = ( gλ /2π) = 5.63 ft/s

Group celerity,Cg = Cp/2 = 2.816ft/s

Wave energy can be calculated according to the linear wave theory

E = 1/8 ρgHm2

= 1/8 * 1.94 *32.174 *32.174 * 0.75*0.75

=141.20 lb/s2

Power = E*Cg

=141.20*2.817

=397.62 lbft/s

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If concentrations of acid and salt in the CH3COOH/CH3COONa buffer solution are equal, pH of the buffer solution is about..?

Solution

Having the equation of buffer:

pH = pKa + log(A-/HA)

but concentrations of acid and salt are equal,

then,log(A-/HA) = 0

We also know that the pKa for CH3COOH = 4.76

Thus pH = 4.76+0 = 4.76

Thus the pH of the buffer solution is about 4.76

Calculate the energy of an electron in the n=1 level of a hydrogen atom. Energy = ______ J

Solution

According to quantum mechanics theory, the energy of an electron of hydrogen atom can be given as:

E=-Eo /n2

Here, n=1,2,3.. and Eo=13.6 ev

substitute, n=1 in the above expression as follows:

E=-13.6/12 ev

=-13.6 ev

Convert electron volt (ev) into joule as follows:

1 ev=1.602x10-19J

-13.6 ev=-13.6x1.602x10-19J

= -21.78x10-19 J

= -2.178x10-18 J

What multiple of Ka must the initial concentration of a weak acid exceed, for the initial concentration and the equilibrium concentration to be within 9.40 percent of each other?  

In an aqueous solution of a weak acid, HA, the assumption is often made that [HA] equilibrium = [HA]initial. This approximation is reasonable if the initial concentration of the acid is sufficiently large compared to Ka. What multiple of Ka must the initial concentration of a weak acid exceed, for the initial concentration and the equilibrium concentration to be within 9.40 percent of each other?

Solution

Let initial concentration [HA]o = x M

 The equilibrium concentration =[HA] = (100 - 9.40)/100 * xM

= 0.906xM

The change in concentration [HA]o - [HA] = x M - 0.906xM = 0.094xM = [A-] = [H+]

The equilibrium constant Ka = [H+] [A-] /HA = 0.094x * 0.094x / 0.906x = 0.00975x

Hence, the initial concentration of a weak acid is x/Ka = x /0.00975x =102.56 =10

Hence, multiple of Ka must the initial concentration of a weak acid exceed, for the initial concentration and the equilibrium concentration to be within 9.40 percent of each other will be = 103 - 1 = 102 multiple of Ka (1 subtracted because of round off above)

Mixed gas Laws- Two liters of oxygen at 25 degree Celsius and a pressure of 2.3 atm is added to a 3.0 L flask containing nitrogen with a pressure of 3.5 atm at 25 degree Celsius. What is the pressure of the gas mixture.

Solution

We can calculate the new pressures for each gas the add to find the pressure of the gas mixture

we know that P1*V1 =P2 *V2 (the temperature remains the same)

For oxygen

P1= 2.3atm

V1=2L

P2=?

V2=3.0L

P2 = 2.3 x 2 /3 = 1.53 atm

Pressure of the mixture

Ptot = PO2 + PN2 = 1.53 + 3.5

= 5.03 atm