Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute.
a. 0.100 m K2S
b. 21.5 g of CuCl2 in 4.50 * 102 g water
c. 5.5% NaNO3 by mass (in water)​

Answers

Answer 1

The freezing points of each of the solutions are as follows;

0.100 m K2S -   - 0.558oC

21.5 g of CuCl2 in 4.50 * 102 g water -   -2oC

5.5% NaNO3 by mass (in water)​ -     - 2.6oC

What is freezing point?

The freezing point is the point at which liquid changes to solid. Let us now look at the freezing point of each solution.

a)

Since;

ΔT = K m i

K = 1.86 oC m-1

m = 0.100 m

i = 3

ΔT =   1.86 oC m-1 * 0.100 m * 3 = 0.558oC

Freezing point = 0oC - 0.558oC =  - 0.558oC

b) Number of moles of CuCl2 = 21.5 g/134.45 g/mol = 0.16 moles

molality = 0.16 moles/0.45 Kg = 0.36 m

ΔT = K m i

ΔT =  1.86 oC m-1 * 0.36 m * 3 = 2oC

Freezing point = 0oC - 2 = -2oC

c) Number of moles of NaNO3 = 5.5g/85 g/mol = 0.065 moles

molality of the solution =  0.065 moles/0.0945 Kg = 0.69 m

ΔT =  1.86 oC m-1 * 0.69 m * 2 = 2.6oC

Freezing point = 0oC - 2.6oC = - 2.6oC

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Related Questions

How are Jupiter, Saturn, and Uranus similiar? How are Jupiter, Saturn, and Uranus similiar?

Question 5 options:

a They are all large, gaseous planetary bodies.


b They are all planets that are the exact same size.


c They are all the same distance from the sun.


d They all have the exact same surface temperature.

Answers

Answer:

a

Explanation:

because they all are gasses

Write a balanced equation for the precipitation reaction that was used to separate Ba2+ from Ca2+ and Mg2+.

Answers

Answer:

[tex]{\rm Ba^{2+}} \, (aq) + {\rm {SO_4}^{2-}}\, (aq) \to {\rm BaSO_{4}}\, (s)[/tex].

Explanation:

Make use of the fact that calcium sulfate [tex]{\rm CaSO_{4}}[/tex] and magnesium sulfate [tex]{\rm MgSO_{4}}[/tex] are much more soluble in water than barium sulfate [tex]{\rm BaSO_{4}}[/tex].

When sulfate ions [tex]{\rm {SO_{4}}^{2-}}[/tex] are added to dilute solutions containing [tex]{\rm Ba^{2+}}\, (aq)[/tex], [tex]{\rm Ca^{2+}}\, (aq)[/tex], and [tex]{\rm Mg^{2+}}\, (aq)[/tex], precipitation would be visible only in the solution with [tex]{\rm Ba^{2+}}\, (aq)\![/tex]. Barium sulfate would be the precipitate.

[tex]{\rm Ba^{2+}} \, (aq) + {\rm {SO_4}^{2-}}\, (aq) \to {\rm BaSO_{4}}\, (s)[/tex].

This ionic equation is balanced as it conserves both the atoms and the charges on the ions.


Determine the hybridization about O in CH3oh

Answers

Answer:

c

Explanation:

lol

I need help to balance a Balance the redox reaction under basic conditions Fe+2 +MnO4-1 ---> Fe+3 + Mn+2

Answers

Answer:

[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} + 8 {H}^{ + } \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]

Explanation:

To balance any redox reaction the primary factor is to identify the species undergoing oxidation and reduction.

Below are the steps to identify species undergoing oxidation and reduction:

Calculate the oxidation number of each atom individually of reactants as well as product.find out the change in oxidation number from reactants to products.if the oxidation number of products are greater than reactants then the species undergone oxidation.if the oxidation number of products are lesser than reactants then the species undergone reduction.

let's calculate oxidation number,

given reaction

[tex]{Fe}^{ + 2} +{MnO_4}^{ - 1} \rightarrow {Fe}^{ + 3} + {Mn}^{ + 2}[/tex]

Of reactant,

[tex]Fe = +2, {MnO_4}^{ - 1} \\ Let \: oxidation \: of \: Mn \: is \: x \: and \: Oxygen -2 \\ {MnO_4}^{ - 1} \rightarrow \: x + 4 \times ( - 2) = - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x + - 8 = - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x = + 8 - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x = + 7 \\ \fbox{ Mn \: +7, O \: -2}[/tex]

Of product,

[tex]\fbox{Fe \: +3, \: Mn \: +2}[/tex]

Now calculate the change in oxidation number

[tex]Fe = +3-(+2) = +1 \\ Mn = +2-(+7) = -5[/tex]

now balance the increase in oxidation number and decrease in oxidation number,

thus we require 5 Fe for 1 MnO4-,

Now putting these numbers as their coefficient,

[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2}[/tex]

now balance oxygen atom on product side by adding water,

since there are 4 oxygen atoms in reactants 4 water molecules will be there on product side.

[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]

now balance Hydrogen atom on reactant side by adding H+ ion, since there are 4 water molecules on product 8 H+ ion will be in reactants!

[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} + 8 {H}^{ + } \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Confirming the balanced reaction, by checking number of atoms and charge on reactant vs Product side.

Total number of Fe on reactant and product = 5

Total number of Mn on reactant and product= 1

Total number of Oxygen on reactant and product = 4

Total number of hydrogen on reactant and product= 8

Total charge on reactant side= 5×(+2)-1+8=17

Total charge on product side = 5×(+3)+1×(+2)=17

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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A student is investigating the affect of different salts on melting points. Four patches of ice of equal size are roped off and a different type of salt is poured on each, one receives table salt (NaCl), one receives Calcium Chloride (CaCl2), one receives Potassium Carbonate (KCO3) and the fourth receives inert sand instead. Each patch receives an equal amount of salt or sand. What could the student investigate?
A. The air temperature near each ice patch.
B. How quickly the ice patches melt.
C. Amount of sand absorbed.
D. The amount of oxygen released as each ice patch melts.

Answers

Answer:

B

Explanation:

By finding how quickly the ice pathes melt you are able to tell how well the salt dissolves in water (ice is the solid form of H2O). Different types of salts will melt the ice at different speeds.

This is also the only answer choice that correlates with melting points so :)

A reversible exothermic synthesis (combination) reaction produces one mole of gaseous product from two moles of gaseous
reactant. Which set of condtions will shift the reaction farthest toward the products?
A)
low temperature, low pressure
B)
low temperature, high pressure
9
high temperature, low pressure
D)
high temperature, high pressure
A 2.0g piece of Mg reacts with HNO3. Which conditions would produce the GREATEST reaction rate?

Answers

Factors that will increase product formation are;

low temperature and high pressure increasing the surface area and concentration

What factors affect chemical equilibrium?

The position of chemical equilibrium is affected by the following factors:

temperature concentration pressure for gases

When equilibrium is altered by an external factor, the reaction shifts to annul the effect of that change.

For the reversible exothermic synthesis, low temperature and high pressure will shift the reaction farthest toward the products.

For the reaction of the 2.0g piece of Mg with HNO3. increasing the surface area of magnesium metal as well as increasing the concentration of HNO3 would produce the GREATEST reaction rate.

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25.88 grams of tin (ll) phosphate reacts with 31.73 grams of zinc.
a write a balanced chemical equation
b. what is the limiting reactant?
c. What is the excess reactant and how many moles of it are left over?
d. How many grams of of the products are produced? (2 answers)​

Answers

Explanation:

Zn + Sn3(PO4)4 = Zn3(PO4)2 + Sn - Balanced Chemical Equation

4.3 is limiting

is the answer

if you like my answer please like comment and mark me as brilliant

Calculate the specific heat in J/(g·ºC) of an unknown substance if a 2.50-g sample releases 12.0 cal as its temperature changes from 25.0ºC to 20.0ºC. _________J/(g·°C)

Answers

Answer:

[tex]\fbox{c = - 4.01 \: joule/g°C}[/tex]

Step by step explanation:

Given:

Mass of given sample (m) = 2.50 g

Initial temperature (T1) = 25°C

Final temperature (T2) = 20°C

Heat Energy Q = 12 cal

To find:

[tex]Specific \: Heat \: c = \: ?[/tex]

Solution:

We know that,

Specific heat of any substance is directly proportional to the mass and change in temperature.

Represented by equation,

[tex]Q = mc \triangle T[/tex]

Where,

Q = Heat Energy

m = mass of given sample

c = specific heat

∆T = change in temperature

Substituting corresponding values,

[tex]Q = mc \triangle T \\ 12 = 2.5\times c \times (20-25) \\ c = \frac{12}{2.5 \times ( - 5)} \\ c = - 0.96 \: cal/g°C \\ [/tex]

We also know that,

[tex]1 \: cal = 4.184 \: joules[/tex]

multiplying above answer by 4.184,

[tex]c = - 0.96 \times 4.184 \\ \fbox{c = - 4.01 \: joule/g°C}[/tex]

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If a car jack's output force is 1000 N and the mechanical advantage is 6.8, what is the input force?

Answers

Answer
147.06N
Explanation:
MA=Load/Effort
6.8=1000/x
Finding x you make x the subject and divide 1000 by 6.8 which is 147.06N

An electromagnet _____________
A. is stronger than a permanent magnet
B. points north
C. can be turned on and off
D. makes noise

I need the answer fast PLS

Answers

I think the answer is A

Answer:

C. can be turned on and off

Explanation:

An electromagnet becomes magnetic when it is part of a complete electric circuit. A switch is a part of an electric circuit that completes or break the circuit turning the circuit on and off.

(*) Sorry for my late answer but I hope this helps others that are looking for this.

I got 100%  ;)

A gas made up of homonuclear diatomic molecules escapes through a pinhole 0.378 times as fast as He gas. Write the chemical formula of the gas.

Answers

For a gas made up of homonuclear diatomic molecules escapes through a pinhole 0.378 times as fast as He gas, the chemical formula of the gas  is mathematically given as

N2(nitrogen gas)

What is the chemical formula of the gas.?

Generally, the equation for the rate to molar relationship  is mathematically given as

\frac{r1}{r2}=\frac{mm2}{mm1}

Thereofore

\frac{r}{r2}=0.378

Hence

0.378=\sqrt{\frac{4.00268}{Mx2}}

Mx2=28.013g/mol

In conclusion,The name of gas with a molar mass of 28.013g/mol is N2

Therefore, the chemical formula of the gas. is N2

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How many grams of air are required to complete the combustion of 186 g of phosphorus (M= 31 g/mol) to diphosphorus pentoxide, assuming the air to be 23% oxygen by mass?
4P(s) + 5O2(g) — 2P2Os(s)

Answers

Answer:

1043.5 g of air  to the nearest tenth.

Explanation:

4P + 5O2 ---> 2P2O5

4*31 g P reacts with 5*32 g oxygen

124 g   ..   ..   .   ....  .   160g oxygen

186 g   ..   ..   ..   ..   ..   (160*186)/124 g oxygen

-  so the mass of air required =  [(160*186)/124] / 0.23

= 240 / 0.23

= 1043.5 g of air.

NEED HELP ASAP!!! As atmospheric carbon dioxide increases, the amount of carbon dioxide dissolved in the ocean _____.

Question 1 options:

increases


decreases


stays the same

Answers

Answer:

If temperature stays the same, the amount of carbon dioxide dissolved in the ocean would increase.

Explanation:

The carbon dioxide in the air ([tex]{\rm CO_{2}}\, (g)[/tex]) and the carbon dioxide dissolved in the ocean ([tex]{\rm CO_{2}}\, (aq)[/tex]) are in a solution equilibrium:

[tex]{\rm CO_{2}}\, (g) \rightleftharpoons {\rm CO_{2}}\, (aq)[/tex].

Assume that the concentration of carbon dioxide in the air ([tex]{\rm CO_{2}}\, (g)[/tex]) increased while temperature stayed the same.

By Le Ch[tex]^{}[/tex]atelier's Principle, the solution equilibrium [tex]{\rm CO_{2}}\, (g) \rightleftharpoons {\rm CO_{2}}\, (aq)[/tex] would shift to offset this increase in [tex]{\rm CO_{2}}\, (g)[/tex] concentration. Specifically, this equilibrium would reduce the amont of [tex]{\rm CO_{2}}\, (g)\![/tex] in the system by converting more atmospheric [tex]{\rm CO_{2}}\, (g)\!\![/tex] to [tex]{\rm CO_{2}}\, (aq)\![/tex] dissolved in the ocean. Therefore, the concentration of [tex]{\rm CO_{2}}\, (aq)\!\![/tex] dissolved in the ocean would increase.

A gas occupies a volume of 50.0 mL at 27°C. At what temperature, in °C, would the pressure be
101.3 kPa if the volume remains constant?

Answers

Answer:

i believe 223

Explanation:

T2 + 500k = 223 C*

Order the atoms from largest to smallest: Cs , Si , Sn .

Answers

Answer:

Si, Sn, Cs

Explanation:

                       Atomic no.          Atomic radius

Silicon Si_______14___________0.117
Tin         Sn______50___________0.140
cesium Cs______55___________0.262

Silicon is the smallest among these

dk how to do this help please

Answers

Explanation:

hello

the answer is b

O or oxygen has 16 molar mass and C or carbon has 12

so CO2 or carbon dioxide=(2×O)+C=(2×16)+12=46

2. A 0.2719 g sample containing CaCO3 reacted with 20.00 mL of 0.2254 M HCl. Given that HCI was excess. The excess HCl required exactly 20.00 mL of 0.1041 M NaOH to reach the end-point using phenolphthalein indicator. Determine percentage purity of CaCO3 in the sample. The reraction involved is CaCO3(s) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l) The titration reaction is HC(aq) + NaOH(aq) NaCl(aq) + H₂O(1)

Answers

The calculated percentage purity of the original sample of calcium carbonate is 44%.

What is excess titration?

In the back titration of the excess acid, we neutralize the excess titrand left in a system.

We have the reaction equation;

CaCO3 + 2HCl ----> CaCl2 + H2O + CO2

To get the moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles

The reaction involving the excess acid occurs thus;

HCl + NaOH ----> NaCl + H2O

Number of moles of NaOH = 0.1041 M * 20/1000 = 0.0021 moles

We have a  1:1 hence 0.0021 moles of HCl reacted also

Number of moles of HCl that reacted with CaCO3 initially = 0.0045 moles - 0.0021 moles = 0.0024 moles

Given the stoichiometry of the reaction,  0.0012 moles of CaCO3 reacted.

Hence;

Mass of pure CaCO3 in the sample= 0.0012 moles * 100 g/mol = 0.12g

Percent purity of the sample = 0.12g/0.2719 g * 100/1

= 44%

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What is the volume of hydrogen chloride under normal conditions, if its mass is 73 g?​

Answers

Explanation:

36.5 g of HCl occupies 2.24×10-2m3 . ∴ 73 g of HCI at STP will occupy 2.24×10-236.5×732=4.48×10-2m3.

Consider the following equilibrium:

N2(g) + 3H2(g) ⇌ 2NH3(g) + 92 kJ

The forward reaction is

Select one:

a.
exothermic and entropy is increasing.


b.
exothermic and entropy is decreasing.


c.
endothermic and entropy is constant.


d.
endothermic and entropy is increasing.


e.
endothermic and entropy is decreasing.

Answers

Answer:

b.exothermic and entropy is decreasing

2. A 0.2719 g sample containing CaCO3 reacted with 20.00 mL of 0.2254 M HCl. Given that HCI was excess. The excess HCl required exactly 20.00 mL of 0.1041 M NaOH to reach the end-point using phenolphthalein indicator. Determine percentage purity of CaCO3 in the sample. The reraction involved is CaCO3(s) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l) The titration reaction is HC(aq) + NaOH(aq) NaCl(aq) + H₂O(1)

Answers

The percentage purity of the calcium carbonate is  44%. The percentage purity gives the amount of pure CaCO3 in the sample.

What is excess titration?

In excess titration or back titration, we neutralize the excess titrand left in a system.

We have the reaction;

CaCO3 + 2HCl ----> CaCl2 + H2O + CO2

Number of moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles

The reaction of the excess acid is according to the reaction;

HCl + NaOH ----> NaCl + H2O

Number of moles of NaOH = 0.1041 M * 20/1000 = 0.0021 moles

Since the reaction is 1:1, 0.0021 moles of HCl reacted also

Number of moles of HCl that reacted with CaCO3  =  0.0045 moles - 0.0021 moles = 0.0024 moles

If 2 moles of HCl reacts with 1 mole of CaCO3

0.0024 moles of HCl reacts with 0.0024 moles * 1/2

= 0.0012 moles

Mass of pure CaCO3 present = 0.0012 moles * 100 g/mol = 0.12g

Percent purity of the sample =  0.12g/0.2719 g * 100/1

= 44%

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How much energy (in J) is required to raise the temperature of 19.2 g of water by 5.00 oC? (Specific heat capacity of liquid water: 4.184 J/g⋅°C)

Answers

Answer: 401.664J

Explanation:
Formula for energy required = mass x specific heat capacity x change in temperature
First of all, check the units and make sure they correspond
Grams to grams and °C to °C
So we can go ahead and multiply
If we were given a parameter in kilogram or kelvin, might've had to convert
Energy required = 19.2 x 4.184 x 5 = 401.664 J
You could approximate to 401.7J

Nitric acid (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas:

2NO(g) + O2(g) → 2NO2(g)

In one experiment, 0.886 mole of NO is mixed with 0.503 mol of O2.

Calculate which of the two reactants is the limiting reagent. (5 points)

Calculate also the number of moles of NO2 produced. (5 points)

What reactant is left over and how much of it is left over? (5 points)

Answers

Answer:

Lim reag = NO        .07 mole of O2 left over

Explanation:

From the equation, you need twice as many moles of NO as O2

  2 x moles O2 =     2 * .503 moles = 1.06 moles of NO needed....you do not have enough so this is the limiting reagent    O2 will be left over

    1/2 * .866 = .433  moles of O2 used    .503 - .433 = .07  mole O2 leftover

Brainliest
Which choice is an example of the use of sound energy in everyday life?


batteries powering toys


fireplaces keeping people warm


lights helping people see


car horns preventing accidents

Answers

pretty sure it's car horns bc it's the only thing that makes noise

Answer:

car hornsss

Explanation:

they loud asl

Which of the following is the correct definition of nuclear energy? A. energy of an object due to the random motion of its atoms and molecules B. energy an object has because of its motion or position C. energy produced from the splitting of atoms D. energy resulting from the flow of charged particles, such as electrons or ions

Answers

Answer:

c. energy from splitting energy

Explanation:

What is ostwald process?? Give the first catalytic reaction of this process. .


Don't spam​

Answers

Answer: Smelling salts and oxygen are introduced into a metal catalyst-containing tube (platinum). Typically warmed to get the reaction started. The alkali is then oxidized to produce nitric oxide.

Answer:

The Ostwald process is a chemical process used for making nitric acid. Wilhelm Ostwald developed the process, and he patented it in 1902. The Ostwald process is a mainstay of the modern chemical industry, and it provides the main raw material for the most common type of fertilizer production

Explanation:

The first stage of the Ostwald process involves the catalytic oxidation of ammonia into nitric oxide, using platinum as the catalyst. The nitric oxide is then transferred to a different oxidizing tower, where it is oxidized into nitrogen dioxide

What effect would you expect this to have on CO2 levels in the atmosphere?

Answers

expect what to have?!?!?!?

where does the beta particle come from

Answers


The beta particle, which may be either negatively charged (negatrons) or positively charged (positrons), originates from the nucleus of an atom. A beta particle is emitted from the nucleus of an atom during radioactive decay. The electron, however, occupies regions outside the nucleus of an atom.

I hope this helps! :)

The elements A and B combine to produce two differenr compounds. A3B and AB2. if 0.18 mol of A3B has a mass of 17.7g and 0.18 mol of Ab2 has a mass of 11.3g calculate the molar masses of both elemts a and b.​

Answers

Taking into account the definition of a system of linear equations and molar mass, the molar masses of elements A and B are 26.778 [tex]\frac{g}{mole}[/tex] and 17.998[tex]\frac{g}{mole}[/tex] respectively.

System of linear equations

A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.

Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values ​​of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.

Definition of molar mass

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.

Molar masses of elements A and B

Being the molar masses of the compounds:

A₃B: [tex]\frac{17.7 g}{0.18 moles} =[/tex] 98.33 [tex]\frac{g}{mole}[/tex]AB₂: [tex]\frac{11.3 g}{0.18 moles} =[/tex] 62.77 [tex]\frac{g}{mole}[/tex]

and considering the definition of molar mass of a compound, the system of equations to be solved is

[tex]\left \{ {{3mA + mB=98.33\frac{g}{mole} } \atop {mA + 2mB=62.77\frac{g}{mole} }} \right.[/tex]

where mA and mB are the molar masses of elements A and B respectively.

There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.

Isolating the variable mB from the first equation:

mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3mA

and substituting in the second equation, you get:

mA + 2× (98.33 [tex]\frac{g}{mole}[/tex] - 3mA)=62.77[tex]\frac{g}{mole}[/tex]

Solving:

mA + 2× 98.33 [tex]\frac{g}{mole}[/tex] - 2×3mA=62.77[tex]\frac{g}{mole}[/tex]

mA + 196.66 [tex]\frac{g}{mole}[/tex] - 6mA=62.77[tex]\frac{g}{mole}[/tex]

mA - 6mA=62.77[tex]\frac{g}{mole}[/tex] - 196.66 [tex]\frac{g}{mole}[/tex]

(-5)mA= - 133.89 [tex]\frac{g}{mole}[/tex]

mA= (- 133.89 [tex]\frac{g}{mole}[/tex] )÷ (-5)

mA= 26.778 [tex]\frac{g}{mole}[/tex]

Remember that mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3mA, replacing the value of mA:

mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3×(26.778 [tex]\frac{g}{mole}[/tex])

mB= 80.334 [tex]\frac{g}{mole}[/tex]

mB= 17.998 [tex]\frac{g}{mole}[/tex]

Finally, the molar masses of elements A and B are 26.778 [tex]\frac{g}{mole}[/tex] and 17.998[tex]\frac{g}{mole}[/tex] respectively.

Learn more about

system of equations:

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molar mass:

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Which object will have the greatest gravitational pull?

Question 5 options:

Venus


Sun


Earth


Moon

Answers

Answer:

the sun i think because the sun has the greatest gravitational pull

Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Cd2+]=0.00190 M and [Au3+]=0.788 M . Standard reduction potentials can be found in this table. 3Cd(s)+2Au3+(aq)↽−−⇀3Cd2+(aq)+2Au(s)

Answers

For a [Cd2+]=0.00190 M and [Au3+]=0.788 M conc, the cell potential  is mathematically given as

Ecell=1.98V

What is the cell potential for the galvanic cell?

Generally, the equation for the  Chemical reaction is mathematically given as

Cd2+(aq) + 2e – → Cd(s)

Au3+(aq) + 3e– → Au(s)

Therefore

Ecell=Ecell-0,0582/nlog

Therefore

Ecell=Ered-Eoxy

Ecell=1.498-(-0.403)

Ecell=1.901

In conclusion

Ecell=1,901-0,0592/6log(0,0016/0.88)^2

Ecell=1.98V

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