The Law of Conservation of Energy states that Energy can neither be ____________ nor _____________ but only change from one __________ to another.

The pH of the standard hydrogen electrode that has electrode potential of -0.128 V  is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log  1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

 -0.128 =  - 0.0296 log 1/[H^+]

-0.128/  - 0.0296  =  log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

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This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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Answer:

where is the question????????????

Answer:

we heat up because the component with lower boiling evaporates first,

leaving the other behind

Answer:   ionosphere

Explanation:       First it bounces off a top layer of the atmosphere called the ionosphere, then it bounces back to the Earth (this is reflection. It then bounces up again to the ionosphere, and continues bouncing back again until it reaches the radio receiver. This is called a skywave, which works around 3 to 30 MHz.

Answer:

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answer: Tristan is most like the Golgi Body

Explanation:

Answer:

The formula for Avagadro's law is V1/n1 = V2/n2, where V = volume and n = amount of gas (in moles).

Explanation:

Answer:

13.6 is the correct answer written in standard form.

Explanation:

1.36, move the decimal once to the right to get 13.6

Answer:

13.6

Explanation:

The standard form is 13.6

Answer:

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.

Answer:

dinitrogen pentachloride

Explanation:

Oils are made up of long hydrocarbon chains which are non polar. Since water is a polar solvent oil doesn't dissolve in water well.( Polar solutes dissolve in polar solvents and non polar solutes dissolve in non polar solvents) The hydrocarbon chains are hydrophobic.

Answer: Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances. These processes are called chemical reactions and, in general, are not reversible except by further chemical reactions.

A physical change is are changes affecting the form of a chemical substance, but not its chemical composition. Physical changes are used to separate mixtures into their component compounds, but can not usually be used to separate compounds into chemical elements or simpler compounds.

Answer:

from the words below underline six example of rhetorical patterns

Answer:

d. 83 °c

Explanation:

Answer:

D. 83 c

Explanation:

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Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.

Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.

Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.

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The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.

The correct answer is sp3.

The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.

The valence electrons are in p orbitals and p orbit mixed after the s orbitals.

Therefore, the correct answer is sp3

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Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:

[tex][HF]=[HF]_0-2x[/tex]

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]

[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]

Finally, the initial concentration of HF is calculated as follows:

[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]

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There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.

NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.

But if  the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.

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A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?

A) Extract with dilute NaOH

B) Extract with dilute HCl

C) Extract with dichloromethane

D) Extract with water

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Answer:

composition of an element

Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.

Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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The original liquid is regarded as a mixture.

A mixture is regarded as a material which comprises of two or more

substances which are combined physically. An example is the mixture of

dye and water.

A compound on the other hand is combined chemically and not physically

which when  exposed to the same type of situation either remains in the

beaker or nothing at all is found. Since the water evaporates and a green

solid was present then it means the material was a solution in which

water(solvent) and the green solid(solute) was physically combined and not

chemically combined together.

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The value of the work done in joule is 405.3 J.

We know that the work done is obtained using the relation;

w = PΔV

Where;

w = work done

P = pressure

V = volume

Now, substituting values,

w = 0.800( 15.0 -  10.0)

w = 4 atm L

Since;

1 L atm = 101.325 J

4 atm L  = 4 atm L × 101.325 J/1 L atm

= 405.3 J

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Answer: Atomic number

Explanation:

I hope this helps you!

Explanation:

Carbon dioxide and water

I hope it helps

Answer:

The usual products of combustion reactions are carbon dioxide and water.

Explanation:

Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

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Answer:

247 ml

Explanation:

How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml

0.000150 x ? = 0.0370 moles HF

? = 0.0370/0.000150 = 247 ml

check

247 ml = 247/1000 L = 0.247

(0.247) x (0.150) =0.370 check

Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].

The balanced reaction is:

H₂(g) +Cl₂(g) ⇆ 2 HCl(g)

Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.

In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is

[tex]K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }[/tex]

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case, the constant Kc can be expressed as:

[tex]K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }[/tex]

You know that in an equilibrium mixture of HCl, Cl₂, and H₂:

Replacing in the expression for Kc:

[tex]4x10^{18} =\frac{[HCl]^{2} }{[1x10^{-11} ]x[2x10^{-10} ] }[/tex]

Solving:

[tex]4x10^{18} =\frac{[HCl]^{2} }{2x10^{-21} }[/tex]

[tex]4x10^{18} x 2x10^{-21}=[HCl]^{2}[/tex]

[tex]8x10^{-3} =[HCl]^{2}[/tex]

[tex]\sqrt[2]{8x10^{-3}} =[HCl][/tex]

0.0894 [tex]\frac{mol}{L}[/tex]= [HCl]

Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].

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