A spontaneous galvanic cell consists of a Pb electrode in a 1.0 M Pb(NO3)2 solution and a Cd electrode in a 1.0 M Cd(NO3)2 solution. What is the standard cell potential for this galvanic cell

Answer:

Answers are in the explanation

Explanation:

It is possible to obtain K of equilibrium of related reactions knowing the laws:

A + B ⇄ C K₁

C ⇄ A + B K = 1 /K₁

The inverse reaction has the inverse K equilibrium

2A + 2B ⇄ 2C K = K₁²

The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients

For the reaction:

2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K

1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)

This is the inverse reaction but also the coefficients are dividing in the half, that means:

[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]

2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)

Here,the only change is the coefficients are the half of the original reaction:

[tex]K_2 = K^{1/2}[/tex]

3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)

This is the inverse reaction. Thus, you have the inverse K of equilibrium:

[tex]K_3 = \frac{1}{K}[/tex]

Answer: The standard cell potential for the cell is +0.51 V

Explanation:

Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]

[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]

The given reaction is:

[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]

As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.

[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]

where both [tex]E^0[/tex]  are standard reduction potentials.

Thus putting the values we get:

[tex]E^0_{cell}=-0.25-(-0.76)[/tex]

[tex]E^0_{cell}=0.51V[/tex]

Thus the standard cell potential for the cell is +0.51 V

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

Explanation:

To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;

Step 1: Divide the percentage composition by the atomic mass

Sulphur = 31.42 / 32 = 0.9819

Oxygen = 31.35 / 16 = 1.9594

Flourine = 37.23 / 19 = 1.9595

Step 2: Divide by the lowest number

Sulphur = 0.9819 / 0.9819 = 1

Oxygen = 1.9594 / 0.9819 ≈ 2

Flourine = 1.9595 / 0.9819 ≈ 2

This means the ratio of the elements is 1 : 2: 2

The empirical formular (simplest formular of a compound) of the compound is;

SO₂F₂

To obtain the molecular formular (Actual formular of a compound);

(SO₂F₂)n = 102.1

Inserting the atomic masses and solving for n;

(102)n = 102.1

n ≈ 1

The molecular formular is; (SO₂F₂)₁ = SO₂F₂

Answer: 1.2 millimoles will be left after 2250 years

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]

b) Amount left after 2250 years

[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]

[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]

[tex]\log\frac{1.6}{a-x}=0.117[/tex]

[tex]\frac{1.6}{a-x}=1.31[/tex]

[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]

Thus 1.2 millimoles will be left after 2250 years

Answer:

im pretty sure its the anode

Explanation:

To solve such, we must know the concept of electrolysis reaction. The correct option is option D among all given options. At anode electrode oxygen forms.

Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.

Electrolysis is the process of passing an electric current through a material to cause a chemical change. A chemical change occurs when a material loses or acquires the electron. To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At anode electrode oxygen forms.

Therefore, the correct option is option D among all given options. At anode electrode oxygen forms.

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Answer:

5.92 g

Explanation:

Convert milliliters to liters.

217 mL = 0.217 L

Since molarity (M) is moles per liter(mol/L), multiply the molarity by the volume to find out how many moles you will need.

0.217 L × 0.246 M = 0.05338 mol

Now, convert the moles to grams using the molar mass.  The molar mass of calcium chloride is 110.98 g/mol.

0.05338 mol × 110.98 g/mol = 5.924 g ≈ 5.92 g

You will need 5.92 g of calcium chloride.

Answer:

Cells are the basic structures of all living organisms. Cells provide structure for the body, take in nutrients from food and carry out important functions. ... These organelles carry out tasks such as making proteins?, processing chemicals and generating energy for the cell

Answer: I absolutely love this question! Biology is so interesting, so I always love to answer the curiosity of others regarding biology, such as that!

Cells are simply the basic structures of all organisms, that are living, of course! Cells provide structure for the body, and they also take in nutrients that your body needs from food and they carry out important functions. These organelles carry out tasks such as making proteins, processing chemicals, and generating energy for the cell. Isn’t that cool?

Hope this helped! <3

Answer:

[tex]pH=10.45[/tex]

Explanation:

Hello,

In this case, for the dissociation of the given base, we have:

[tex]base\rightleftharpoons OH^-+CA[/tex]

Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:

[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]

And in terms of the reaction extent [tex]x[/tex] we can write:

[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]

For which the roots are:

[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]

For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:

[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]

And the pH:

[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]

Regards.

The pH of the solution is 10.45.

Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;

              :B(aq) + H2O(l) ⇄ BH(aq)  + OH^-(aq)

I            0.05                        0                0

C           -x                            +x                +x

E        0.05 - x                      x                  x

We know that the Kb of codeine is 1.6 x 10^-6, Hence;

1.6 x 10^-6 = x^2/0.05 - x

1.6 x 10^-6 (0.05 - x ) =  x^2

8 x 10^-8 - 1.6 x 10^-6x =  x^2

x^2 +  1.6 x 10^-6x - 8 x 10^-8 = 0

x = 0.00028 M

The concentration of hydroxide ions = 0.00028 M

Given that pOH = - log[0.00028 M]

pOH = 3.55

pH + pOH = 14

pH = 14 - 3.55

pH = 10.45

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Answer:

Diferentes tipos de arcilla

ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...

ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...

ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...

ARCILLAS “BALL CLAY” O DE BOLA. ...

CAOLIN. ...

ARCILLA REFRACTARIA. ...

BENTONITA.

Explanation:

Answer:

Coil method and the slab method.

Explanation:

Answer:

[tex]MM_{acid}=140.1g/mol[/tex]

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

[tex]n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}[/tex]

Thus, solving for the moles of the acid, we obtain:

[tex]n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol[/tex]

Then, by using the mass of the acid, we compute its molar mass:

[tex]MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol[/tex]

Regards.

Answer:

0.52 g of chromium(II) hydroxide, Cr(OH)2.

Explanation:

We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.

This can be obtained as follow:

Molarity of CrCl2 = 0.258 M

Volume = 35.9 mL = 35.9/1000 = 0.0359 L

Mole of CrCl2 =?

Molarity = mole /Volume

0.258 = mole of CrCl2 /0.0359

Cross multiply

Mole of CrCl2 = 0.258 x 0.0359

Mole of CrCl2 = 0.0093 mole

Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.

This can be obtained as follow:

Molarity of KOH = 0.338 M

Volume = 35.8 mL = 35.8/1000 = 0.0358 L

Mole of KOH =.?

Molarity = mole /Volume

0.338 = mole of KOH /0.0358

Cross multiply

Mole of KOH = 0.338 x 0.0358

Mole of KOH = 0.0121 mole.

Next, we shall write the balanced equation for the reaction. This is given below:

2KOH + CrCl2 → Cr(OH)2 + 2KCl

From the balanced equation above,

2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

2 mole of KOH reacted with 1 mole of CrCl2.

Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.

From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.

Therefore, KOH is the limiting reactant.

Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.

In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.

The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:

From the balanced equation above,

2 mole of KOH reacted to produce 1 mole of Cr(OH)2.

Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.

Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.

This is illustrated below:

Mole of Cr(OH)2 = 0.00605 mole

Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol

Mass of Cr(OH)2 =..?

Mole = mass /Molar mass

0.00605 = mass of Cr(OH)2/86

Cross multiply

Mass of Cr(OH)2 = 0.00605 x 86

Mass of Cr(OH)2 = 0.52 g

Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.

Answer:

[tex]128~ATP[/tex]

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:

[tex]Number~of~Rounds=\frac{n}{2}-1[/tex]

Where "n" is the number of carbons, in this case "18", so:

[tex]Number~of~Rounds=\frac{18}{2}-1~=~8[/tex]

We also have to calculate the amount of Acetyl-Coa produced:

[tex]Number~of~Acetyl-Coa=\frac{18}{2}~=~9[/tex]

Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex]. So, if we have 8 rounds we will have 8 [tex]FADH_2[/tex] and 8 [tex]NADH[/tex].

Finally, for the total calculation of ATP. We have to remember the yield for each compound:

-) [tex]1~FADH_2~=~2~ATP[/tex]

-) [tex]1~NADH~=~3~ATP[/tex]

-) [tex]Acetyl~CoA~=~10~ATP[/tex]

Now we can do the total calculation:

[tex](8*2)~+~(8*3)~+~(9*10)=130~ATP[/tex]

We have to subtract  "2 ATP" molecules that correspond to the activation of the fatty acid, so:

[tex]130-2=128~ATP[/tex]

In total, we will have 128 ATP.

I hope it helps!

Answer:

[tex]pH=2.28[/tex]

Explanation:

Hello,

In this case, for the acid dissociation of formic acid (HCOOH) we have:

[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]

Whose equilibrium expression is:

[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]

That in terms of the reaction extent is:

[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]

Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:

[tex]x=0.00528M[/tex]

[tex][H^+]=0.00528M[/tex]

Then, as the pH is computed as:

[tex]pH=-log([H^+])[/tex]

The pH turns out:

[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]

Regards.

Answer:

ΔG = - 31.7kJ/mol

Explanation:

It is possible to find ΔG of a reaction at certain temperature knowing Kc following the equation:

ΔG = ΔG° + RT ln Q

ΔG° = -RT lnKc

ΔG = -RT lnKc + RT ln Q (1)

Where R is gas constant (8.314J/molK), T absolute temperature (350°C + 273.15 = 623.15K) and Q reaction quotient

For the reaction,

3H₂(g) + N₂(g) ⇄ 2NH₃(g)

Q = [NH₃]² / [H₂]³[N₂]

Where the concentrations of each chemical are:

[NH₃] = 1.0mol / 2.5L = 0.4M

[H₂] = 5.0mol / 2.5L = 2M

[N₂] = 2.5mol / 2.5L = 1M}

Q = [0.4M]² / [2M]³[1M]

Q = 0.02

And replacing in (1):

ΔG = -RT lnKc + RT ln Q

ΔG = -8.314J/molK*623.15K ln 9 + 8.314J/molK*623.15K ln 0.02

ΔG = - 31651J/mol

Answer:

Explanation:

The ionic lattices would have the highest melting point Potassium oxide. option A is correct.

An ionic compound is a giant structure of ions. The ions have a regular, repeating arrangement called an ionic lattice. The lattice is formed because the ions attract each other and form a regular pattern with oppositely charged ions next to each other.

Ionic compounds are held together by electrostatic forces between oppositely charged ions.

These forces are usually referred to as the ionic lattice contains such a large number of ions, that a lot of energy is needed to overcome this ionic bonding so ionic compounds have high melting and boiling points.

therefore, sodium oxide has the highest melting point. option A is correct

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Answer:

To prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should parboil them to cook them to partial doneness.

Answer:

The liquid that is often used in thermometers is chrome.

It is khwon for raising its volule when the temperature raises and vice-versa. ● the temperature and the volume are proprtional to each other so using Mathematics, scientists have figured out a way to benefit from it to make a thermometer.

Answer:

Any ionic molecule formed of a base and an acid, which dissolves in water to produce ions is known as a salt. The four common types of salts are:  

1. NaCl or sodium chloride is the most common kind of salt known. It is also known as table salt.  

2. K2Cr2O7 or potassium dichromate refers to an orange-colored salt formed of chromium, potassium, and oxygen. It is toxic to humans and is also an oxidizer, which is a fire hazard.  

3. CaCl2 or calcium chloride looks like table salt due to its white color. It is broadly used to withdraw ice from roads. It is hygroscopic.  

4. NaHSO4 or sodium bisulfate produces from hydrogen, sodium, oxygen, and sulfur. It is also known as dry acid. It has commercial applications like reducing the pH of swimming pools and spas and others.  

Answer:

The salt breaks into positive chlorine ions and negative sodium icons

Explanation:

The question requested for the wrong option in the list. If we look at the option selected, we will notice that sodium ions are positively charged ions since sodium is a metal. Metals produce cations (positive ions) because they loose electrons. Therefore, a sodium ion can never be negatively charged.

Similarly, chlorine is a highly electronegative nonmetal. It gains electrons in an ionic bond. Hence chlorine ions can not be positive.

Answer:

7.50 L

Explanation:

The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 6.25 atm × 1.20 L / 1.00 atm

V₂ = 7.50 L

Answer:

The correct answer is 2.2 mL.

Explanation:

Given:

Average: 2.9 mL

SD: 0.71 mL

We can define a 1 SD range in which the value of volume (in mL) will be comprised:

Volume (mL) = Average ± SD = (2.9 ± 0.7) mL

Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL

Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL

Thus, the minimum value within a 1 SD range of the average is 2.2 mL

The minimum value within 1 SD is 2.19 mL

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 2.9 mL, σ = 0.71 mL; hence:

The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)

Therefore the minimum value within 1 SD is 2.19 mL

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Answer:

Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.

Answer:

– 844 kJ/mol.

Explanation:

The following data were obtained from the question:

N2(g) + 3 F2(g) –––> 2 NF3(g)

Enthalpy of N≡N (N2) = 945 kJ/mol

Enthalpy of F–F (F2) = 155 kJ/mol

Enthalpy of N–F3 (NF3) = 283 kJ/mol

Enthalpy change (∆H) =?

Next, we shall determine the enthalpy of reactant.

This is illustrated below:

Enthalpy of reactant (Hr) = 945 + 3(155)

Enthalpy of reactant (Hr) = 945 + 465

Enthalpy of reactant (Hr) = 1410 kJ/mol

Next, we shall determine the enthalpy of the product.

This is illustrated below:

Enthalpy of product (Hp) = 2 x 283

Enthalpy of product (Hp) = 566 kJ/mol

Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:

Enthalpy of reactant (Hr) = 1410 kJ/mol

Enthalpy of product (Hp) = 566 kJ/mol

Enthalpy change (∆H) =?

Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)

Enthalpy change (∆H) = 566 – 1410

Enthalpy change (∆H) = – 844 kJ/mol

Answer:

– 844 kJ/mol.

Explanation:

The following data were obtained from the question:

N2(g) + 3 F2(g) –––> 2 NF3(g)

Enthalpy of N≡N (N2) = 945 kJ/mol

Enthalpy of F–F (F2) = 155 kJ/mol

Enthalpy of N–F3 (NF3) = 283 kJ/mol

Enthalpy change (∆H) =?

Next, we shall determine the enthalpy of reactant.

This is illustrated below:

Enthalpy of reactant (Hr) = 945 + 3(155)

Enthalpy of reactant (Hr) = 945 + 465

Enthalpy of reactant (Hr) = 1410 kJ/mol

Next, we shall determine the enthalpy of the product.

This is illustrated below:

Enthalpy of product (Hp) = 2 x 283

Enthalpy of product (Hp) = 566 kJ/mol

Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:

Enthalpy of reactant (Hr) = 1410 kJ/mol

Enthalpy of product (Hp) = 566 kJ/mol

Enthalpy change (∆H) =?

Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)

Enthalpy change (∆H) = 566 – 1410

Enthalpy change (∆H) = – 844 kJ/mol

Explanation:

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

Where X is reaction coordinate

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

Answer:

the volume of the titrant used

Explanation:

Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.

Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).

Answer:

The ΔG° is 29 kJ and the reaction is favored towards reactant.

Explanation:

Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,  

ΔG° = ΔH°rxn - TΔS°rxn

= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K

= 41.2 kJ - 12.2 kJ

= 29 kJ

As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.  

Answer:

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Explanation:

Given:

Change in heat (ΔH) = 150 joules

Temperature (T) = 150 K

Find:

ΔS surrounding (entropy change of the reservoir)

Computation:

ΔS surrounding (entropy change of the reservoir) = - ΔH / T

ΔS surrounding (entropy change of the reservoir) = - 150 / 150

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Answer:

Explanation:

Fe2+ Has 4 unpaired electrons.

By method of elimination;

Option A: Ni2+ has two unpaired electrons. so this option is wrong.

Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.

Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.

Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.

Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.