PLLLSS HURRY IM BEING TIMED

Given the E0 values of the following two half-reactions: Zn  Zn2+ + 2e- E0 = 0.763 volt Fe  Fe2+ + 2e- E0 = 0.441 volt a) Writ

34kurt

Answer:

For a: The balanced chemical equation is written below.

For b: The corrosion of iron pipe will take place in the presence of zinc.

For c: Zinc will not protect iron pipe from corrosion.

Explanation:

For a:

The given half reaction follows:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, zinc will undergo reduction reaction will get reduced.

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V$

Reduction half reaction: $Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V$

The balanced chemical equation follows:

$Fe+Zn^{2+}\rightarrow Fe^{2+}+Zn$

For b:

For a reaction to be spontaneous (thermodynamically feasible) , the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=-0.441-(-0.763)=0.322V$

As, the EMF is coming out to be positive, the reaction will be thermodynamically feasible and corrosion of iron pipe will take place in the presence of zinc.

For c:

As, the EMF of the cell is positive, the zinc will not protect the iron pipe from corrosion and the reaction will take place.

4 0

3 years ago

For what order reaction does the half-life get longer as the initial concentration increases?

nasty-shy [4]

The half-life gets longer as the initial concentration increases in zero-order reaction.

The amount of time it takes for the concentration of a given reactant to reach 50% of its initial concentration is known as the half-life of a chemical reaction (i.e. the time taken for the reactant concentration to reach half of its initial value).

For zero order reaction:

The half-life is given as:

$t_{1/2}=\frac{[A]_{0}}{2k}$ where k is the rate constant of the reaction and $[A]_{0}$ is the initial concentration.

As we can see that the half-life is directly proportional to the initial concentration. Therefore, when the initial concentration increases the half-life gets longer.

For the first-order reaction,

The half-life is given as:

$t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{k}$

A first-order reaction's half-life is independent of the initial concentration.

For a second-order reaction,

The half-life is:

$t_{\frac{1}{2} }=\frac{1}{k[A]_{0}}$

The initial concentration is inversely proportional to the half-life, so when the initial concentration increases the half-life will get shorter.

Learn more about half life here:

brainly.com/question/1160651

#SPJ4

3 0

1 year ago

The solubility of lead(ii) chloride is 0.45 g/100 ml of solution. what is the ksp of pbcl2? 4.9 × 10-2 1.7 × 10-5 8.5 × 10-6 4.2

Artyom0805 [142]

Answer:
1.7 * 10^-5

Explanation:
1- get the number of moles of PbCl2:
number of moles = mass / molar mass
number of moles = 0.45 / 278.1 = 1.618 * 10^-3 moles

2- get the concentration of Pb2+:
molarity = number of moles of solute / volume of solution in liters
molarity = (1.618 * 10^-3) / (0.1) = 0.0162 M

3- getting concentration of Cl-:
PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)
We can note that:
For a certain amount of Pb2+ formed, twice this amount of Cl- is formed.
This means that:
for 0.0162 M of Pb2+, 2*0.0168 = 0.0324 M of Cl- is formed

4- getting Ksp:
Ksp = [Pb2+][Cl-]²
Ksp = (0.0162)*(0.0324)²
Ksp = 1.7 * 10^-5

Hope this helps :)

7 0

4 years ago

Throughout the reflection, make sure you have a copy of the Student Guide and your data tables. Use the drop-

Inessa05 [86]

Answer:

1) mass and type of material

2) type of material

3) temperature

Explanation:

8 0

2 years ago

Sort each of the following events based on whether the solubility of the indicated gas will increase, decrease, or stay the same

STatiana [176]

The answers are: 1) decrease 2) increase 3) decrease 4) increase 5) increase

<h3>FURTHER EXPLANATION</h3>

The following factors affect the solubility of gases in liquids:

1. Temperature - inverse relationship

As the temperature is increased, the gas particles gain energy and start to move faster. This enables them to escape the liquid leading to decrease in the solubility of the gas.

2. Partial Pressure of gas - direct relationship

According to Henry's Law, for a given temperature, the solubility of the gas and the partial pressure of the gas above the liquid are proportional. With the application of pressure on the gas above the liquid, the gas particles are forced to move into the solvent.

1. Temperature is increased - gas solubility decreases (Factor 1)

2. Partial pressure of an anesthetic gas is increased - gas solubility increases (Factor 2)

3. O2: temperature of a body of water rises - gas solubility decreases (Factor 1)

4. Pressure of a gas over a solvent is increased - gas solubility increases (Factor 2)

5. Air in blood: a diver descends 10 m and the pressure increase by 1 atm - gas solubility increases (Factor 2)

As the diver goes deeper, the pressure increases and more gases dissolve in the diver's blood. As the diver ascends, he or she must do so slowly to ensure that this large amount of gas is removed from the blood gradually otherwise, the diver might suffer from decompression sickness.

<h3>LEARN MORE</h3>

Learn more about Charles' Law brainly.com/question/7259721

Learn more about Ideal Gas Law brainly.com/question/4147359

Learn More about Boyle's Law brainly.com/question/8511562

Keywords: solubility, Henry's Law

6 0

4 years ago

Read 2 more answers