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## If 0.499 moles of zinc reacts with excess lead (iv) sulfate, how many grams of zinc sulfate would be produced in the following reaction? Pb(SO4)2 + 2Zn ——–>2ZnSO4 + Pb

Given ;

moles of Zn = 0.499

having the balanced equation

Pb(SO4)2 + 2Zn ——————-> 2ZnSO4 + Pb

From the equation, 2 moles of Zinc reacts to produce 2 moles of Zinc sulfate

0.499 moles of Zn will produce 2/2 x 0.499 = 0.499 moles of ZnSO4

moles of ZnSO4 produced = 0.499 moles

we know that molar mass of ZnSO4 = 161.47 g/mol

mass = molar mass x moles

mass = 161.47g/mol x 0.499 mol

=80.57 g

Therefore 80.57g of zinc sulfate will be produced

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## A student wanted to prepare 500 mL 0.20 mol/L NaOH solution for an acid-base titration lab. If 0.50 mol/L NaOH is the only source, how many mL of 0.50 mol/L NaOH should he use to do the 500-mL 0.20 mol/L NaOH solutuion?

From the principle of dilution,M1V1 = M2V2,where M1 is the concentration in molarity (moles/Liters) of the concentrated solution, V2 is the volume of the concentrated solution, M2 is the concentration in molarity of the dilute solution (after more solvent has been added), and V2 is the volume of the dilute solution.

M1=0.50 mol/L

V1=?

M2=0.20 mol/L

V2=500mL

V1 =M2V2/M1

V1= (0.2 x 500)/0.5 = 200

Therefore 200mL of 0.50 mol/L NaOH should be used.

## Describe how a buffer resists changes in pH explain how to apply the Henderson-Hasselbach equation to solve for the pH of a buffered solution.

We know that a buffer is a solution that can resist pH change upon the addition of an acidic or basic components.The resisting of pH change is made possible by the ability of the buffer solution to neutralize small amounts of acid or base that are added into a solution, this therefore maintains the pH of the solution relatively stable.

• Henderson-Hasselbach equation can be used to solve for pH for both acidic and basic buffers
• For acidic buffers: pH = pKa + log10([A-]/[HA]) where Ka is the dissociation constant for the weak acid, [A-] is the concentration of conjugate base and [HA] is the concentration of the weak acid.
• For basic or alkaline buffers, the Henderson-Hasselbach equation : pH = 14 – (pKb + log10([B+]/[BOH])) where Kb is the dissociation constant for the weak base, [B+] is the concentration of conjugate acid and [BOH] is the concentration of the weak base.

## 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 the 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

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

Solution

Thus the pH of the buffer solution is about 4.76

Explanation:

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=1.602×10-19J

-13.6 ev=-13.6×1.602×10-19J

= -21.78×10-19 J

= -2.178×10-18 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)