105/22 • (1.251 - 0.620)=

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:

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:

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:

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:

[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]

Explanation:

You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.

1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]

2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].

The moles of a compound has been given as:

[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]

The moles in 23.45 g copper (II) hydroxide has been:

[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]

The moles of copper (II) hydroxide has been 0.24 mol.

The number of molecules in 0.24 mol sample has been driven by:

[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

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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:

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 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:

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:

27°C or 300K

Explanation

We were told that the pressureof the system decreased by 10 times implies that P2= P1/10

Where P2=final pressure

P1= initial pressure

Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1

Where T2= final temperature =-123C= 273+(-123C)=150K

T1= initial temperature

But from gas law

PV=nRT

As n and R are constant

P1V1/T1 = P2V2/T2

T1= P1V1T2/P2V2

T1=2×T2

T1=2×150

T1=300K

=300-273

=27°C

the initial temperature (°C) of a system is 27°C

Answer:

Molar solubility of AgBr = 51.33 × 10⁻¹³

Explanation:

Given:

Amount of NaBr = 0.150 M

Ksp (AgBr) = 7.7 × 10⁻¹³

Find:

Molar solubility of AgBr

Computation:

Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr

Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150

Molar solubility of AgBr = 51.33 × 10⁻¹³

When The Molar solubility of AgBr is = 51.33 × 10⁻¹³

Given as per question:

The Amount of NaBr is = 0.150 M

Then Ksp (AgBr) is = 7.7 × 10⁻¹³

Now we Find:

The Molar solubility of AgBr

The we Computation is:

The Molar solubility of AgBr is = Ksp (AgBr) / Amount of NaBr

After that Molar solubility of AgBr is = 7.7 × 10⁻¹³ / 0.150

Therefore, Molar solubility of AgBr is = 51.33 × 10⁻¹³

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Answer:

2.3 × 10⁻⁹

Explanation:

Step 1: Write the reaction for the solution of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 2: Make an ICE chart

We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.

        CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

I                                0                 0

C                              +S               +S

E                                S                 S

The solubility product constant is:

Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹

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:I think it's Geosphere

Explanation:

Answer:

A

Explanation:

Geo means rock, or earth. Hydro means water, Atmosphere is space, and Bio global ecosystem composed of living organisms

Answer:

The empirical formulae for the car fuel is C4H9

Answer:

Zinc- anode

Copper- cathode

Sodium sulphate- salt bridge

Explanation:

A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.

In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e

Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)

Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.

Answer:

zinc=anode

copper=cathode

Explanation:

Answer:

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

Explanation:

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:

Organic chemist? I do not know.

Explanation:

Thanks you.

The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.

The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.

Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.

Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.

According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.

Thus, option C is correct.

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Answer:

550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M

Explanation:

It is possible to find the pH of a buffer by using H-H equation:

pH = pKa + log [A⁻]/[HA]

For the formic buffer (HCOOH/HCOONa):

pH = 3.74 + log [HCOONa]/[HCOOH]

As you need a buffer of pH 3.65:

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

Where  [HCOONa]/[HCOOH] can be taken as the moles of each specie.

As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:

Replacing (2) in (1):

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

And moles of HCOONa are:

[HCOONa] = 0.1 mol - 0.055mol =

As concentration of the solutions is 0.1M, the volume you need to add of both solutions is:

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.

Here we used the H-H equation:

pH = pKa + log [A⁻]/[HA]

Now

For the formic buffer (HCOOH/HCOONa):

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

Now

need a buffer of pH 3.65:

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

0.81283 = [HCOONa]/[HCOOH] (1)

here  [HCOONa]/[HCOOH] can be considered as the moles of each specie.

Now the total moles should be

0.10 moles = [HCOONa] + [HCOOH] (2)

Now

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

[HCOOH] = 0.055 moles

And moles of HCOONa should be

[HCOONa] = 0.1 mol - 0.055mol =

[HCOONa] = 0.045 moles

Now

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

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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:

[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:

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:

[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:

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:

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.

Learn more about Avogadro's number,here:

https://brainly.com/question/11907018

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Answer:

310 mL

Explanation:

Step 1: Given data

Step 2: Calculate the initial volume

We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.400 M × 5.50 L / 7.10 M

V₁ = 0.310 L = 310 mL

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:

[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