If 100-mL of 1.0 M Sr(OH)2 is added to 100 mL of 1.0 M HCl, the pH of the mixture would be _____. Group of answer choices

Answer:

what does little birdie say in the birth of their differences lak lak lak nu pasand aayi baby sleep are no longer children all strong industry all strong baby to show the meaning of rice is here to get up from sleep meaning of lips is Hasan let the mother is saying the baby to sleep in a new

Taking into account the definition of molarity and the molar mass of the compound, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M  solution is 31 g.

In first place, you have to know tha molarity is a measure of the concentration of that substance that indicates the number of moles of solute present in the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.

[tex]molarity=\frac{number of moles of solute}{volume of solution}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

In this case, you know:

So, by definition of molarity, the number of moles is calculated as:

[tex]0.90 M=\frac{number of moles of solute}{0.125 L}[/tex]

Solving:

number of moles of solute= 0.90 M× 0.125 L

number of moles of solute= 0.1125 moles

On the other side, molar mass is the mass of one mole of a substance, which can be an element or a compound. In this case, the molar mass of FeSO₄.7H₂O is 277.85 [tex]\frac{g}{mole}[/tex].

Then you can apply the following rule of three: if by definition of molar mass, 1 mole of the compound contains 277.85 g, 0.1125 mole contains how much mass?

[tex]mass=\frac{0.1125 moles*277.85 g}{1 mole}[/tex]

Solving:

mass ≅ 31 g

Finally, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M  solution is 31 g.

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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:

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:

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:

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:

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: 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:

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:

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:

– 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:

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:

[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:

Corrective lense or just lens.

Explanation:

Answer:

c

Explanation:

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

2

5

C) 2 Is the correct answer, I took the test and it was correct.

According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.

Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.

It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .

According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.

Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2

Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.

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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]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:

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:

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:

Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.

Explanation:

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:

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 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: Gamma Radiation

Explanation:

The emission of Gamma rays does not cause a change in both the atomic and mass number. They are  electromagnetic radiation.

The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.

Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.

The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.

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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.

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:

[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

Answer:

a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.

b. Therefore electrons flow from the aluminium electrode to the nickel electrode.

c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.

Explanation:

Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.

Given:

E ⁰N i ⁺² = − 0.23 V   is the standard reduction potential for the nickel ion

E ⁰  A l ⁺³ =  − 1.66  V  is the standard reduction potential for the aluminum ion

The most negative potentials  correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.

So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.

The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.

Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.

Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.

So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.

Answer:

E) NaF and SrO

Explanation:

The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.

In which pair do both compounds exhibit predominantly ionic bonding?

A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.

B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.

C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.

D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.

E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.

Answer:

ethyl 4-bromobenzoate

Explanation:

In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):

[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]

This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.

Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.

[tex]O=C-O-CH_2-CH_3[/tex]

Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.

See figures 1 and 2 to further explanations.

I hope it helps!

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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