A metal object is to be gold-plated by an electrolytic procedure using aqueous AuCl3 electrolyte. How much gold may be deposited in 3.0 min by a constant current of 10. A

The concentration of hydroxide ion of a solution with a pH of 8. 25 is 1.77×[tex]10^{6}[/tex] M

Calculation,

The pH of a solution tells us about concentration of the [tex]H^{+[/tex] ion ( proton ) in solution. However, if we know the value of pH , we can calculate pOH with the help of pH and from pOH we can find concentration of hydroxide ions in the solution.

pH + pOH = 14

8. 25 + pOH = 14

pOH = 14 - 8. 25 = 5.75

Since, pH = -㏒[ [tex]H^{+[/tex] ]

Similarly, pOH = -㏒[[tex]OH^{-[/tex] ]

If pOH = 5.75 , that means that -㏒[[tex]OH^{-[/tex] ]  = 5.75 ,which means that,

[[tex]OH^{-[/tex] ]  = [tex]10^{-5.75}[/tex] = 1.77×[tex]10^{6}[/tex].

So, we know that the concentration of the [tex]OH^{-[/tex] ions in the solution with pH of 8.25 is 1.77×[tex]10^{6}[/tex] Molar

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Answer:

subscript

Explanation:

H₂ O   means there are two hydrogen atoms and one oxygen atom

               the ' ₂ '  is a subscript

The  Cm of the Microsand concentration in the tanks when expressed in grams of Microsand per liter (g/L) is known to be 3.8 g/L.

This is known to be a kind of  aggregate, that is said to be exempted from clay and shale and it is one that can rightly fine to pass via a No. 100 (that is 150 µm) sieve.

Note that from the question:

The Average of six samples = 30mg/L.

Then one need to Multiply:

30 by 300gpm x 1 Train x 1700 ( this is the use of the bulk density conversion factor).

Then one need to also Divide by (4000 gpm x 2000 mL) =  1.8 g/L.

Lastly you then multiply by 2,

1.8 g/L x 2  = 3.8 g/L.

Therefore, looking at the solution above, the Cm of the Microsand concentration in the tanks  is seen to be 3.8 g/L.

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x2=x+2 at x=−1 and x=2 so we have no need to worry about the end-points

f(x)=x+2−x^2

df/dx=1–2x

and that is zero (indicating a maximum) at x=1/2

So the maximum distance is f(1/2)=2.5–0.25=2.25

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The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.

The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.

It can be found using the formula,

m = Z I t

where, m = mass of metal deposited = 75.8g

            Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001

             I is the current passed

              t is the time taken = 75hour = 75 × 60 = 4500s

On subsituting in above formula,

75.8 = E I t / F

⇒ 75.8 = 0.0001 × I × 4500

⇒ I = 168.4 Ampere (A)

Hence, amount of current required to deposit a metal is 168.4A.

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If the volume of the container were increased to 4.0 L, more N₂ and H₂ would be produced (Option D)

This is simply defined as a state in a chemical system where there is no observable change in the properties of the system with time.

A French scientist postulated a principle which helps us to understand a chemical system in equilibrium.

The principle states as follow:

If a an external constraint such as change in temperature, pressure or concentration is imposed on a system in equilibrium, the equilibrium will shift so as to neutralize the effect.

From the principle given above, we can see that volume does not affect equilirium position.

However, from Boyle's law, we understood that pressure and volume are in invest relationship.

Thus, increasing volume simply means decreasing pressure.

A decrease in pressure will favors the side where there is an increase in volume.

Let us consider the equation from the question:

N₂(g) + 3H₂(g) <=> 2NH₃(g)

Thus, we can conclude that increasing the volume (i.e decreasing the pressure) of the reaction will favors the backward reaction, hence, more N₂ and H₂ would be produced.

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option (B) KOH is the correct answer

KOH is the substance that, when added to the solution, is most likely to raise the pH .

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The answer is A.

We know the formula for molarity :

Molarity = Number of moles ÷ Volume

Let's assume there are 0.6 moles of NaCl in 1 liter of ocean water.

Then, mass of salt is :

Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).

What is standard entropy?

The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."

Calculation:

Balancing the given reaction following-

1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)

ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]

Here S° = standard entropy of the system

Insert into the aforementioned equation all the typical entropy values found in the literature:

ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]

⇒ΔS° = - 99.4 J/K

Therefore, the standard entropy, ΔS° is -99.4 J/K.

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Epimerization and Enediol rearrangement are the two most common unwanted side reactions which can occur when monosaccharides are treated with base.

Epimerization is a process in stereochemistry in which there is a change in the configuration of only one chiral centre. As a result, a diastereomer is formed. The classical example of this in medicine is tetracycline. In acidic conditions around pH 4, tetracycline readily undergoes epimerization at position 4, and an inactive 4-epitetracycline is produced, which on dehydration forms 4-epianhydrotetracycyline, a highly toxic product. This toxic compound can also be formed from acid catalysed (at lower pH) dehydration of tetracycline via anhydrotetracycline.

Enediol rearrangement is a  transformation which occurs at basic medium and allows the conversion of epimers, defined as isomeric forms that differ in the position of the hydroxyl group at C-2. In this way it is possible to transform through the enediol intermediate glucose to mannose and vice versa.

Another important isomerization process through the enediol rearrangement is the interconversion of glucose and fructose. Thus, the enolization proceeds by migration of proton at position 2, to carbon at 1.

Monosaccharides contain both alcohol and carbonyl functional groups. This allows monosaccharides to undergo many of the reactions typical for these functional groups. In particular, alcohols can be converted to esters, converted to ethers, converted to acetals, or oxidized. Carbonyls can be reacted with nucleophiles, be reduced to form alcohols, or be oxidized to form carboxylic acids.

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The enol carbon or ∝-carbon  nucleophile attacks at molecular bromine in the acid-catalyzed α-bromination of a ketone

Treatment of ketones with bromine in the presence of acid will results in formation of a new C-Br bond at the alpha position. The purpose of the acid  is to catalysed formation of the enol from ketone , which is active nucleophile in the reaction. This reaction is called haloform reaction which is used to identify the methyl substituted ketone in the presence of aldehyde.

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Answer: Everything can dissolve in water because water is considered a universal solvent

Explanation: Liquid water is cooler than solid salt and sugar, allowing them to dissolve easily.

According to the concept of solubility, as water is considered to be a universal solvent salt and sugar both are able to dissolve in water as water is a universal solvent.

Solubility is defined as the ability of a substance which is basically solute to form a solution with another substance. There is an extent to which a substance is soluble in a particular solvent. This is generally measured as the concentration of a solute present in a saturated solution.

The solubility mainly depends on the composition of solute and solvent ,its pH and presence of other dissolved substance. It is also dependent on temperature and pressure which is maintained.Concept of solubility is not valid for chemical reactions which are irreversible. The dependency of solubility on various factors is due to interactions between the particles, molecule or ions.

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Rusting of Iron is the formation of rust which is characterized by formation of layer of red colour. Formula of Rust is Ferric Oxide (Fe₂O₃) where iron exhibits oxidation state of +3.

Following are the indicators of rusting:

Even if the Reddish brown rust is not visible on the iron surface, then also Ferroxyl Indicator turns blue in presence of Iron ions.

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Explanation:

(e)the heat is absorbed into the reaction .

Answer:

Density = 0.27 decagrams/cm³

Explanation:

From conversion tables, we know that;

1 g/cm³ = 0.1 decagrams/cm³

We are given;

Density of aluminium = 2.7 g/cm³

Thus;

Density = 2.7 * 0.1

Density = 0.27 decagrams/cm³

Answer:

270000 decagrams/m³

Explanation:

1.

Density=mass/Volume

=2.7g/1cm3

=(2.7/1000)/(1/1000000)

=2.7x1000

Density=2700kg/ m3

= 270000 decagrams/m³

2.

1000g=1kg

1g=1/1000kg

1cm3= ? m3

100cm=1m

1cm=1/100 m

1cm3=1/1000000 m3

A conjugate acid is given by the acid-base theory of Bronsted–Lowry. The conjugate acid of a weak base, trimethylamine is ((CH₃)₃NH⁺) trimethylammonium ion.

A conjugate acid is a compound that has been formed when a base accepts the hydrogen or the proton ion from an acid. It can also be said that a hydrogen ion is added to a base.

The conjugate acid differs from the base by the addition of one proton ion to it. The reaction of a weak base, trimethylamine can be shown as,

(CH₃)₃N(aq) + H₃O⁺(aq) ⇌ (CH₃)₃NH⁺(aq) + H₂O(l)

Here, the trimethylamine compound has accepted a proton from hydronium to produce a conjugate acid, trimethylammonium cation, (CH₃)₃NH⁺.

Therefore, trimethylammonium (CH₃)₃NH⁺ is the conjugate acid of trimethylamine.

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The ph of the solution will be acidic. It is because of the higher ka value of ammonia as compared to kb of nitrate

As we know, A salt solution contains ammonium cation (ka = 5.7 × 10^–10) and nitrite anion (kb = 1.4 × 10^–11) in equal proportions but the ka and kb are different.

As we see that the value of ka is higher than the kb that's why the ph level will be acidic.

Acidic solutions are sour in taste as compared to alkaline solutions.

We know that the ph scale basically ranges from 0 to 14. The ph solutions less than 7 are acidic and greater than 7 are basic.

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27.94 g/mole is the molar mass of the unknown gas.

Use Graham's Law, which indicates that the square root of a gas's molar mass determines its rate of effusion.

The following equation compares two distinct gases, (a) and (b), and expresses this:

sqrt(M(b)/M) = R(a)/R(b) (a)

R is the effusion rate.

square root = sqrt

the molar mass, M

In response to your query, if gas (a) is the unknown gas and Xe(M = 131 g/mol) is the second gas, we have:

sqrt(M(Xe)/M(a) = 2.165, where R(a)/R(Xe)

Square both sides, and you get:

(2.165)² = 131 g/mol/ M (a)

M(a)(2.165)² = 131

M(a) is equal to 131/(2.165)² = 27.94 g/mol.

Having a molar mass of 27.94 g/mol, the unknown gas.

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A single slit with a width of 2019 * 109 m creates its initial minimal for 624 nm light at an angle of 18°.

How does diffraction work?

Waves spreading outward around obstructions are known as diffraction. Diffraction happens with sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as with incredibly minuscule moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities. Diffraction prevents the creation of sharp shadows as one of its effects. In order to spread out and illuminate regions at which a shadow is anticipated, light must be bent around corners, which is known as diffraction.

Calculation:

Provided for a single slit, m=1

λ = 624 *10⁻⁹

sinθ = sin 18⁰

Therefore,

asinθ=mλ

a = [tex]\frac{1 * 624 *10^{-9} }{sin 18}[/tex]

⇒a = 2019 *10⁻⁹ m

Therefore the width of a single slit is 2019 *10⁻⁹ m.

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Na -k pumps couple atp hydrolysis with the transport of Na and k.

Both ions are used for active transport.

d. K+ and Na+ both diffuse into the cell along their concentration gradients and drive the transport of glucose.

Na/K pump is a pump located on the plasma membrane which uses ATP to move 3 Na ions out the cell and brings in 2 K ions into the cell. It is an example of primary active transport. As a consequence, concentration of Na is higher outside the cell, while K concentration is higher inside the cell.

Glucose is transported in the cell against its gradient, together with Na ions (symport) which move down their concentration gradient.

This is an example of secondary active transport because it uses the energy from the primary active transport to move other substances such as glucose against their own gradients.

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At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid.  HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R   [tex]HSO_4^-[/tex]    ⇄  [tex]H^+ + SO_4^2^-[/tex]

I    [tex]0.14[/tex]

C   [tex]- x[/tex]               [tex]+x[/tex]       [tex]+x[/tex]

E   [tex]0.14-x[/tex]        [tex]x[/tex]         [tex]x[/tex]

[tex]K_a = 1.3[/tex] × [tex]10^-^2[/tex] for [tex]HSO^-_4[/tex] . As a result,

[tex]\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a[/tex]

[tex]K_a[/tex] is large. It is no longer valid to approximate that [tex][HSO^-_4][/tex] at equilibrium is the same as its initial value.

[tex]\frac{x^2}{0.14-y} = 1.3 * 10^-^2[/tex]

[tex]x^2+1.3*10^-^2x - 0.14[/tex] × [tex]1.3[/tex] × [tex]10^-^2= 0[/tex]

Solving the quadratic equation for [tex]x , x \geq 0[/tex] since [tex]x[/tex] represents a concentration;

                             [tex]x=0.0366538[/tex]

Then, round the results to 2 significant figure;

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Fluids used for an intravenous fluid transfusion must be isosmotic with bodily fluids.

Intravenous fluids, sometimes abbreviated as "IV fluids," are liquids that are administered to replace water, sugar, and salt that you may need if you're ill or having surgery and are unable to eat or drink regularly. Through a drip, Intravenous fluids are administered directly into a vein.

In order to treat or prevent dehydration, Intravenous fluids are carefully prepared liquids that are injected into a vein. They are applied to patients of all ages who are ill, hurt, becoming dehydrated from physical activity or the heat, or who are having surgery. Rehydrating intravenously is a straightforward, risk-free treatment that is frequently used.

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The volume in the final system that was initially enclosed in a 12.3 liter cylinder with a moveable piston at 300 K and 2.0 atm is 36.9L.

The volume of a given amount of gas can be calculated by using the combined gas law formula as follows;

P1V1/T1 = P2V2/T2

Where;

According to this question, 1.0 mole of a gas is enclosed in a 12.3 liter cylinder with a moveable piston at 300 K and 2.0 atm.

However, half of the gas is removed, leaving 0.50 mole in the cylinder and the system is warmed to 900 K. The volume in the final system can be calculated as follows:

2 × 12.3/300 = 2 × V2/900

0.082 = 2V2/900

73.8 = 2V2

V2 = 73.8 ÷ 2

V2 = 36.9L

Therefore, the volume in the final system that was initially enclosed in a 12.3 liter cylinder with a moveable piston at 300 K and 2.0 atm is 36.9L.

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Answer:

Explanation:

it's 18.45, my sister just got the guestion right! hope it helps :)

Correct option is C

Latent heat of vaporization is the amount of heat required to bring about the phase transition from liquid to gaseous state, at its boiling point.

The substance is Propane :

10140000 J of heat is used to boil 28.47 kg at 298 K. Here, the important thing is, the liquid is changing to vapor and the temperature is not changing means it's a phase change. So, the formula used for this is:

                        q=m× Δ[tex]H_{vap}[/tex]

where, q is the heat energy, m is mass and  Δ[tex]H_{vapour}[/tex]  is the enthalpy of vaporization.

Let's rearrange the formula for  as:

Given mass of the liquid = 28.47kg

Heat required to boil the liquid = 10.14 ×[tex]10^{2} J[/tex]

Let's plug in the values: [tex]\frac{10.14*10^{2}J }{28.47Kg} =356164\frac{J}{Kg}[/tex]

Rounding off 356164 will turn into 356000 J[tex]Kg^{-1}[/tex]

Hence, the right choice is C. propane 356000 j[tex]Kg^{-1}[/tex]

What is latent heat ?

Latent heat, energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

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A symmetrical acid anhydride will undergo a reaction with an alcohol to give an ester and a carboxylic acid. An asymmetrical acid anhydride will undergo a reaction with an alcohol to give two esters and two carboxylic acids.

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In a solution, when the concentrations of a weak acid and its conjugate base are equal: 1. the -log of the concentration of H+ and the -log of the Ka are equal.

Any acid that entirely ionises in solutions is considered a strong acid. When positioned, it emits the most hydrogen ions or protons.

Acids that don't entirely dissociate in solution are referred to as weak acids. In other words, any acid that is not a strong acid qualifies as a weak acid. The amount of dissociation determines how strong an acid is; the more dissociation, the stronger the acid.

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Urea get decomposed during autoclaving, because it is volatile. It is filter sterilized and added aseptically to your autoclave media.

Can urea be autoclaved?

What is autoclave sterilization?

What is filter sterilization?

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Answer:

1 mole of H2SO4 contains 2 x 6.023 x 10²³ So, 0.1 mole of H2SO4 contains =2 x 6.023 x 10²³ x 0.1

= 1.2 x 10²³atoms of hydrogen

hello there !!

In the Question it's asked that total atoms present in 0.1 mol of H2So4.

so , in the question number of moles is given as 0.1 mole

=> n = 0.1 mole

and total atoms present are = ???

So , in H2So4 molecule there are two hydrogen atoms

=> H2So4 → 2 hydrogen atoms

one mole of H2So4 consist of ;

=> 2 × 6.022 × 10²³

1 mol = 6.022 × 10 ²³ atoms

Therefore, 0.1 mole of H2So4

=> 0.1 × 2 × 6.022 × 10²³ Atoms

=> 0.1 × 2 × 6.022 × 10²³ Atoms=> 12.004 × 10²² Atoms