Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE

Answer:

C

Explanation:

Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.

Answer:

Rubidium

Explanation:

Answer:

The correct answer is 2.75 grams of HCl.

Explanation:

The given balanced equation is:  

CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams

The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,  

= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.  

Answer:

2 moles of HNO3

Explanation:

The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.

If that is the case then 2 moles of HNO3 will be produced.

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

Answer:

[tex]\boxed{From \ +6 \ to \ 0}[/tex]

Explanation:

2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)

In the given reaction, Iron in the reactants side have the oxidation number of +6. This is because [tex]O_{3}[/tex] with [tex]Fe_{2}O_{3}[/tex] has oxidation state -6, So any atom with it would have an oxidation state of +6 to give the resultant of zero.

In the products side, Iron acts as a free element reacting with no other atom. So, as per the rule of oxidation states, the oxidation state of Iron in the products side will be zero.

So, the oxidation number changes from +6 to 0 .

Extra Info: Decrease in oxidation state is Reduction , So Iron is being reduced here.

The change in the oxidation number of the iron atom in the reaction is from +3 to 0

Oxidation is simply defined as the the loss of electron. However, Oxidation number simply talks about the number of electrons that is either gained or lossed during bond formation.

The change in the oxidation number of iron in the reaction can be obtained as follow:

2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g)

Oxidation number of Fe₂O₃ = 0 (ground state)

Oxidation number of oxygen = –2

Fe₂O₃ = 0

2Fe + 3O = 0

2Fe + 3(–2) = 0

2Fe – 6 = 0

Collect like term

2Fe = 6

Divide both side by 2

Fe = 6/2

Thus, the oxidation number of Fe in Fe₂O₃ is +3

Therefore, the change in the oxidation number of the iron, Fe, atom in the reaction is from +3 to 0

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Answer:

See explanation

Explanation:

For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion [tex]H^+[/tex]).

With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.

In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.

See figure 1

I hope it helps!

Answer:

1. b.) The wavelength gets shorter as the frequency increases  

2. X-rays < Ultraviolet radiation < Visible light < Infrared radiation

3. 2 × 10⁵ J

Explanation:

1. Wavelength vs frequency

fλ= c

f = c/λ

Thus, frequency and wavelength are inversely proportional.

The wavelength increases (gets longer) as the frequency decreases.

2. Order of increasing wavelength

X-rays < Ultraviolet radiation < Visible light < Infrared radiation  

3. Energy of green light

(a) Energy of 1 photon

λ = 5 × 10² nm = 5  × 10² × 10⁻⁹ m = 5 × 10⁻⁷ m

fλ = c

f = c/λ = (2.998 × 10⁸ m·s⁻¹)/(5 × 10⁻⁷ m) = 6 × 10¹⁴ s⁻¹

E = hf = 6.626 × 10⁻³⁴ J·s × 6 × 10¹⁴ s⁻¹ = 4 × 10⁻¹⁹ J

(b) Energy  of 1.0 mol of photons

[tex]\text{Energy} = \text{1.0 mol photons} \times \dfrac{6.022 \times 10^{23}\text{ photons }}{\text{1 mol photons }} \times \dfrac{4 \times 10^{-19}\text{ J}}{\text{1 photon }} = \mathbf{2 \times 10^{5}} \textbf{ J}\\\\\text{The energy of 1.0 mol of photons of green light is $\large \boxed{\mathbf{2 \times 10^{5}}\textbf{ J}}$}[/tex]

Note: The answer can have only one significant figure because that is all you gave for the wavelength of the light.

The representative elements are the elements in the tall columns of the periodic table.

The representative elements that can also be referred to as the main group elements. They can be used to represent the chemistry of the group to which they belong.

Hence, the representative elements are the elements in the tall columns of the periodic table.

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#SPJ1

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

.02M

CH₃COOH  = CH₃COO⁻ + H⁺

C                       xC             xC

Ka = xC . xC / C = x² C

1.8 x 10⁻⁵ = x² . .02

x² = 9 x 10⁻⁴

x = 3 x 10⁻²

= .03

concentration of H⁺ = xC = .03 . .02

= 6 x 10⁻⁴ M , volume =  45 x 10⁻³ L

moles of H⁺  = 6 X 10⁻⁴  x 45 x 10⁻³

= 270 x 10⁻⁷ moles

= 2.7 x 10⁻⁵ moles

concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L

moles of Na OH = 2 X 10⁻²  x 35 x 10⁻³

= 70 x 10⁻⁵ moles

=  

NaOH is a strong base so it will dissociate fully .

there will be neutralisation reaction between the two .

Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles

= 67.3 x 10⁻⁵ moles of NaOH

Total volume = 45 + 35 = 80 x 10⁻³

concentration of NaOH after neutralisation.= 67.3  x 10⁻⁵ / 80 x 10⁻³ moles / L

= 8.4125  x 10⁻³ moles / L

OH⁻ = 8.4125  x 10⁻³

H⁺ = 10⁻¹⁴ / 8.4125  x 10⁻³

= 1.1887 x 10⁻¹²

pH = - log (  1.1887 x 10⁻¹² )

= 12 - log 1.1887

= 12 - .075

= 11.925 .

Answer:

(a) The diode voltage,  [tex]V_D =[/tex]  0.776 V

(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A

Explanation:

Given;

saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A

nonideality factor, n = 1.05

(a) the diode voltage

Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A

Diode voltage is calculated as;

[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]

Where;

[tex]V_T[/tex] is thermal voltage at 25°C = 0.025

[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]

b) the diode current for VD = 0.1 mV

[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]

Answer:

Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Look up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].

The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.

How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?

[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].

There are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

The molarity of a solution measures its molar concentration. For this solution:

[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

(Rounded to four significant figures.)

The given question is incomplete. The complete question is :

Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?​

a) [tex]O^{2-}[/tex]

b)  [tex]F^{-}[/tex]

c)  [tex]N^{3-}[/tex]

d)  [tex]S^{2-}[/tex]

Answer: b)  [tex]F^{-}[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here potassium is having an oxidation state of +1 called as  cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to  give neutral ionic compound.

Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]

Answer:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

[tex]Entalpy=-2861.9~KJ[/tex]

Explanation:

In this case, we have to start with the reagents:

[tex]Al~+~NH_4NO_3[/tex]

The compounds given by the problem are:

-) Nitrogen gas =  [tex]N_2[/tex]

-) Water vapor  =  [tex]H_2O[/tex]

-) Aluminum oxide =  [tex]Al_2O_3[/tex]

Now, we can put the products in the reaction:

[tex]Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_) [/tex]

When we balance the reaction we will obtain:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

Now, for the enthalpy change, we have to find the standard enthalpy values:

[tex]Al_(_S_)=0~KJ/mol[/tex]

[tex]NH_4NO_3_(_a_q_)=-132.0~KJ/mol[/tex]

[tex]N_2_(_g_)=0~KJ/mol[/tex]

[tex]H_2O_(_g_)=~-~241.8~KJ/mol[/tex]

[tex]Al_2O_3_(_S_)=~-~1675.7~KJ/mol[/tex]

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the reagents:

[tex](0*2)~+~(-132*3)=~-396~KJ[/tex]

And the products:

[tex](0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ[/tex]

Finally, for the total enthalpy we have to subtract products by reagents :

[tex](-3125.9~KJ)-(-396~KJ)=-2729.9~KJ[/tex]

I hope it helps!

Answer:

Silver ion - Lewis acid, Ammonia -  Lewis base

Explanation:

The reaction is given as;

Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)]2+(aq)

A lewis acid is an electron pair acceptor. While a lewis base is any substance that that can donate a pair of nonbonding electrons.

This reaction however is a complexation reaction, where ammonia is reacting with the silver ion.

Silver ion accepts electrons in this reaction, hence it is the lewis acid. The ammonia on the other hand donates the electrons used in bonding so it is the lewis base.

Answer: The energy of a Br–F bond is 110 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]

[tex]B.E_{Br-F}=110kJ/mol[/tex]

Thus the energy, in kJ/mol, of a Br–F bond is 110

The correct answer is C. There is no oxygen in space, so there can be no combustion.

Explanation:

Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.

This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).

Answer:

C.) There is no oxygen in space, so there can be no combustion.

Explanation:

I got it correct on founders edtell

Answer:

NH4Br + AgNO3 —> AgBr + NH4NO3

Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

Answer:

103.62 g of AgCl.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2

Step 2:

Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.

This is illustrated below:

Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol

Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g

Thus, from the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Step 3:

Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.

The mass of AgCl produced from the reaction can be obtained as follow:

Form the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.

Therefore, 103.62 g of AgCl were produced from the reaction.

✔ When air molecules collide with things around us, it produces pressing force , which is measured with a Pressure gauge.

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

simple distillation can be used when the temperature difference between the boiling points of two miscible liquid is at least 25°c. the temperature difference between the boiling points of kerosene and petrol is 25c. hence, this mixture can separated using simple distillation.

simple distillation

Answer:

[tex]N=9.033\times 10^{22}\ \text{units}[/tex]

Explanation:

It is given that,

Number of moles, n = 0.15

We need to find the number of particles in 0.15 mol of NaCl. Let N are the number of particles i.e.

Number of particles = number of moles × Avagadro's number

[tex]N=n\times N_A[/tex]

So,

[tex]N=0.15\times 6.022\times 10^{23}\\\\N=9.033\times 10^{22}\ \text{units}[/tex]

So, the number of particles are [tex]9.033\times 10^{22}\ \text{units}[/tex].

Answer:

This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.

Explanation:

In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.

Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.

So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.

Answer:

The Empirical Formular is given as; Ti₆Al₄V

Explanation:

The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.

Elements                        Titanium            Aluminium        Vanadium

Percentage                    64.39                 24.19                   11.42

Divide all through by their molar mass

                                     64.39 / 47.87      24.19 / 27               11.42 / 50.94

                                       =  1.345                = 0.896                 = 0.224

Divide all though  by the smallest number (0.224)

                                     1.345 / 0.224        0.896 / 0.224       0.224 / 0.224

                                     = 6                         = 4                             = 1

The Empirical Formular is given as; Ti₆Al₄V

Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

Titanium :

Divide by Molar mass : = 64.39/47.87 = 1.345

Aluminum :

Divide by Molar mass : = 24.19/27 = 0.896

Vanadium :

Divide by Molar mass : = 11.42/50.94 = 0.224

Divide by the smallest :

Titanium = 1.345 / 0.224 = 6.00

Aluminum = 0.896 / 0.224 = 4

Vanadium = 0.224 / 0.224 = 1

Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

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Answer:

188

Explanation:

For every 2 molecules of methanol reacted, 4 molecules of water are formed.  Use this relationship to solve.

2/4 = 94/x

2x = 376

x = 188

188 molecules of water will be formed.

Answer:

Your answer will either be 22.9897 or 22.990 !!

Explanation:

Answer:

Empirical formula: C₅H₅O

Molecular formula: C₁₀H₁₀O₂

Explanation:

When a compound containing C, H and O elements is combusted, the general reaction is:

CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O

Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.

Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =

2.0x10⁻³ moles of CO₂ = moles C

Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =

1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H

The mass of the moles of C and H are:

2x10⁻³ moles C ₓ (12g / mol) = 0.024g C

2x10⁻³ moles H ₓ (1g / mol) = 0.002g H

Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O

Moles are:

0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O

Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:

C: 2.0x10⁻³ / 4x10⁻⁴ = 5

H: 2.0x10⁻³ / 4x10⁻⁴ = 5

O:  4x10⁻⁴ / 4x10⁻⁴ = 1

Thus, empirical formula is:

The molar mass of the empirical formula is:

12×5 + 1×5 + 16×1 = 81g/mol

As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:

Answer:

C. NO2 + O3 → NO3 + O2 (slow)

NO3 + NO2 → N2O5 (fast)

Explanation:

A reaction mechanism represents an amount of elementary steps that explain how a reaction proceeds. The mechanism must explain the experimental rate law. Also, the slow step is the rate determining step.

This rate law is obtained from the multiplication of the reactants in the slow step, thus:

A. NO2 + NO2 → N2O2 (fast)

N2O4 + O3 → N2O5 + O2 (slow)

Rate law:

rate = k [N2O4] [O3]

This mechanism is not consistent with rate law.

B. NO2 + O3 → NO5 (fast)

NO5 + NO5 → N2O5 + (5/2)O2 (slow)

Rate law:

rate = k [NO5]²

This mechanism is not consistent with rate law.

C. NO2 + O3 → NO3 + O2 (slow)

NO3 + NO2 → N2O5 (fast)

Rate law:

rate = k [NO2] [O3]

D. NO2 + NO2 → N2O2 + O2 (slow)

N2O2 + O3 → N2O5 (fast)

Rate law:

rate = k [NO2]²

This mechanism is not consistent with rate law.

Thus, right solution is:

C. NO2 + O3 → NO3 + O2 (slow)

NO3 + NO2 → N2O5 (fast)