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Answer:

1. REDOX

2. None of the above

3. Precipitation

4. Preicipitation

5. Acid base neutralization

Explanation:

Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.

Option 4 and 3.

3) CaO (s) + CO₂ (g) →  CaCO₃(s)

4) KOH (aq)  + AgCl (aq)  →  KCl (aq)  + AgOH(s)

Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.

5) Ba(OH)₂ (aq)  +  2HNO₂(aq)  →  Ba(NO₂)₂ (aq) + 2H₂O(l)

Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.

1) Ba(ClO₃)₂ (s)  → BaCl₂ (s) +  3O₂(g)

Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).

2.  2NaCl(aq)  +  K₂S(aq)  Na₂S (aq)  + 2KCl (aq)

None of the above

Answer:

4NH₃(g)  + 5O₂(g)  →  4NO(g)  +  6H₂O

2NO(g) + O₂(g) → 2 NO₂

Explanation:

First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.

3H₂ (g) +  N₂(g)  →  2NH₃ (g)

Afterwards, ammonia reacts to oxygen, to produce NO and H₂O

The equation for the process will be:

4NH₃(g)  + 5O₂(g)  →  4NO(g)  +  6H₂O

Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:

2NO(g) + O₂(g) → 2 NO₂

3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)

Answer:

The correct option is d

Explanation:

Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.

A heterogeneous mixture​

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

Answer:

0.3 mol

Explanation:

Assuming HCl is in excess and Zn is the limiting reagent,

from the balanced equation, we can see the mole ratio of Zn:H2 = 1:1,

which means, each mole of zinc reacted gives 1 mole of H2.

So, if 0.30 mol Zn is added, the no. of moles of H2 produced will also be 0.3 mol, since the ratio is 1:1.

Answer:

Explanation:

Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ +  2Na NO₃

Converting into ions

Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻

Cancelling out common terms

Mn⁺ + 2 OH⁻ = Mn( OH)₂

this is net ionic equation required.

Answer:

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Explanation:

The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:

[tex]\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)[/tex]

Where:

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]c(t)[/tex] - Concentration of the compound as a function of time.

The solution of the differential equation is:

[tex]c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]c_{o}[/tex] is the initial concentration of the compound.

The time is now cleared in the result obtained previously:

[tex]\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}[/tex]

[tex]t = -\tau \cdot \ln \frac{c(t)}{c_{o}}[/tex]

Time constant as a function of half-life is:

[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the composite decomposition, measured in seconds.

If [tex]t_{1/2} = 8\,s[/tex], then:

[tex]\tau = \frac{8\,s}{\ln 2}[/tex]

[tex]\tau \approx 11.542\,s[/tex]

And lastly, given that [tex]\frac{c(t)}{c_{o}} = \frac{1}{9}[/tex] and [tex]\tau \approx 11.542\,s[/tex], the time taken for the concentration to decrease to one-ninth of its initial value is:

[tex]t = -(11.542\,s)\cdot \ln\frac{1}{9}[/tex]

[tex]t \approx 25.360\,s[/tex]

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Answer:

b. Binary compounds of carbon and hydrogen

Explanation:

Before proceeding, Hydrocarbons refers to organic chemical compounds composed exclusively of hydrogen and carbon atoms. This means the only elements present in an hydrocarbon are;

- Carbon

- Hydrogen

Looking through the options;

- Option A: This is wrong because the hydroxyl group contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option B: This is correct. Binary compounds refers to compounds with just two elements.

- option C: This is wrong because water contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option D: Carbon atoms can contain other elements so this option is wrong.

- option E: This also wrong because we had already gotten the correct option.

Answer:

% yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Explanation:

given

volume of acetone= 43.8 mL

molar weight of acetone = 58.08 g/mol

density of acetone = 0.791 g/mL

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)

43.8 mL = 43.8mL × 0.791g/mL

= 34.6458g ≈34.65g

1 mole of acetone = 58.08g

∴34.65g = 34.65g/58.08g

= 0.60mol

molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol

mole = mass/ molar weight

mole = 79.4g/ 234.29g/mol

mole(n) = 0.3389mol ≈ 0.34mol

1 mole of acetone will produce 1 mole of the product

∴0.60mol of acetone will produce 0.60mol of the product

but we get 0.34mol of the product

∴ % yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

Answer:

Option A

Explanation:

Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.

Answer:

A. Silicon (Si)

Explanation:

Silicon (Si) is primarily used as a semiconductor material to make electronic chips.

Answer:

0.0913 M

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, we obtained the following:

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 2

Data obtained from the question include:

Molarity of base, NaOH (Mb) = 0.0990 M

Volume of base, NaOH (Vb) = 20.52 mL

Volume of acid, H2C3H2O4 (Va) = 11.13 mL

Molarity of acid, H2C3H2O4 (Ma) =..?

The molarity of the acid, H2C3H2O4 can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 11.13 / 0.0990 x 20.52 = 1/2

Cross multiply

Ma x 11.13 x 2 = 0.0990 x 20.52 x 1

Divide both side by 11.13 x 2

Ma = (0.0990 x 20.52)/ (11.13 x 2)

Ma = 0.0913 M

Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

Effusion rate of [tex]H_2[/tex],

Molar mass of hydrogen,

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

Answer:

Mass PbCl₂ = 50.24g

Mass AgCl = 14.84g

Explanation:

The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:

Ag⁺ + Cl⁻ → AgCl(s)

Pb²⁺ + 2Cl⁻ → PbCl₂(s)

If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:

Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂

And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:

(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl

Moles of Cl⁻ that were added in the KCl solution are:

0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.

Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)

0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles

0.45409 + 0.00021X = 0.46464

0.00021X = 0.01055

X = 0.01055 / 0.00021

X = 50.24g

As X = Mass PbCl₂

And mass of AgCl = 65.08 - 50.24

The masses of the compounds in the precipitate can be found my knowing

the number of moles of chloride ion contributed by each compound.

Reasons:

The given parameter are;

Volume of KCl solution added = 0.242 L

Concentration of KCl solution = 1.92 M KCl

The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions

Precipitates formed = AgCl and PbCl₂

The mass of the precipitate = 65.08 g

Required:

The mass of PbCl₂ and AgCl in the precipitate

Solution;

Number of moles of chloride ions in a mole of PbCl₂ = 2 moles

Number of moles of chloride ions in a mole of AgCl = 1 mole

Let X represent the mass of PbCl₂ in the precipitate, we have;

The mass of AgCl in the precipitate = 65.08 g - X

[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]

Number of moles of chloride ions from PbCl₂ is therefore;

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]

The number of moles of chloride ions from one mole of KCl = 1 mole

Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;

0.242 L × 1.92 moles/L = 0.46464 moles

Number of moles of chloride ions from KCl = 0.46464 moles

[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]

Which gives;

[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]

Therefore;

[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]

The mass of PbCl₂ in the precipitate, X ≈ 49.24 g

The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g

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Answer:

The number of oxide ions as the nearest neighbors of  [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six

Explanation:

The regularity of a crystal structure leads to the idea of space lattice.In order to explain this concept, let us consider a crystal of NaCl, It consists of a perfectly regular arrangement of sodium ions and chlorine ions.

If we represent the position of each Na+ in the crystal by a point marked x the result will be a regular three dimensional network of points. This will be the space lattice of Na+ in the crystal NaCl. The symmetry of the combined lattice determined the symmetry of the crystal as a whole.

The space lattice of a crystal may be considered as built up of a three dimensional basic pattern  called unit cell. The unit cell is a repeat unit which generates the whole pattern in three dimensions of the unit cell.

In Solid MgO , the crystal structure which is used to predict the properties of the material, have the same structure as that of NaCl.

The obtain the structure of a face centered cubic FCC unit cell where the ions occupy the corner of the cube and the center of each face of the cube.

The number of oxide ions as the nearest neighbors of  [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six. As a result of that , the coordination number of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions is six.

Answer:

It will take 6486.92 minutes  for [H2O2] to fall from 0.95 M to 0.33 M

Explanation:

The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.

In first order reaction;

[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]

where;

a = concentration at time t

[tex]a_o[/tex] = initial concentration

and k = constant.

[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]

[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]

[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]

t = 6486.92 minutes

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6

Answer:

a. increase

Explanation:

Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.

By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.

Answer:

PhCH2CH2COOH

Explanation:

This is a reaction of PhCH2CH2Br with KCN in the presence of H3O^+. The reaction first leads to the formation of PhCH2CH2CN.

We must recall that part of the properties of nitriles is that they can be converted to carboxylic acids in the presence of H3O^+. This is a common synthetic route for carboxylic acids.

Therefore, when the PhCH2CH2CN is now further reacted with H3O^+, the carboxylic acid PhCH2CH2COOH is formed as the major organic product of the reaction, hence the answer given above.

Answer:

Graphite> sodium chloride> solid ammonia

Explanation:

Melting points of solids has a lot to do with the nature of intermolecular forces in the solid. A substance melts when the intermolecular forces holding the crystal lattice has been overcome such that that the crystal structure of the solid just collapses.

Graphite consists of covalently bonded layers of carbon atom which form a giant lattice. The melting point of graphite is very high because of the fact that the strong covalent bonds that hold the carbon atoms together in the layers require a lot of heat energy to break. Grapoghite melts at about 3600°C

Sodium chloride is an ionic compound that melts at about 801°C. The lattice is composed of alternate sodium and chloride ions.

Solid ammonia is held together by much weaker intermolecular interaction hence it has a melting point of about −77.73 °C.

Explanation:

Speed = 2345 ÷ 35 = 67m/s

Answer:

Osmolarity of solution of KCI = 0.40 osmol

Explanation:

Given:

KCL ⇒ K⁺ + Cl⁻

Find:

Osmolarity of solution of KCI

When M = 0.20 M

Computation:

1 mole of KCL = 2 osmol

1 M of KCl = 2 Osmolarity

So,

Osmolarity of solution of KCI = 2 × 0.20

Osmolarity of solution of KCI = 0.40 osmol

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

Answer:

495nm

Explanation:

The energy of a photon could be obtained by using:

E = hc / λ

Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.

The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:

242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.

Replacing in the equation:

E = hc / λ

4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ

λ = 4.946x10⁻⁷m

Is maximum wavelength  of light that could break a Cl-Cl bond.

Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:

4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =

Answer:

An alkyl halide can undergo SN2 reaction with an amine

Explanation:

The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.

The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).

This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.

The detailed mechanism of this reaction has been attached to this answer.

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.

[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]

This value indicates that

A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is [tex]ONO^-[/tex]

D. The conjugate acid of CN- is HCN

Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When [tex]K_{p}>1[/tex]; the reaction is product favoured.

When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.

[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.

As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]

Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.

Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]