Read the chemical equation. Mg + 2HCl → MgCl2 + H2 How many moles of MgCl2 are produced from 1 mole of HCl? 0.2 0.5 1.0 1.5

Answer:

The rate at which ZnCl2 appears increases.

Explanation:

Hello,

In this case, the reaction is:

[tex]Zn(s) + 2HCl (aq) \rightarrow ZnCl_2 (aq) + H_2 (g)[/tex]

Therefore, the law of rate proportions is:

[tex]\frac{1}{-1}r_{Zn}= \frac{1}{-2}r_{HCl}= \frac{1}{1}r_{ZnCl_2}= \frac{1}{1}r_{H_2}}[/tex]

In such a way, since the concentration of hydrochloric acid is increasing The rate at which ZnCl2 appears increases, because the addition of a reactant is directly related with the products formation due to the fact that more reactant will yield more product.

Best regards.

Answer:

68%

Explanation:

Since we need a percentage we can use any number we want for our initial value.

5(1/2)^900/1620 = 3.40

(3.40 / 5)*100 = 68%

To make sure lets use a different initial amount

1(1/2)^900/1620 = 0.68

(0.68/1) * 100 = 68%

To solve this question, we'll assume the initial amount of radium-226 to be 1.

Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:

Determination of the number of half-lives that has elapsed.

Half-life (t½) = 1620 years

Time (t) = 900 years

[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]

Determination of the amount remaining

Initial amount (N₀) = 1

Number of half-lives (n) = 5/9

[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]

Determination of the percentage remaining.

Initial amount (N₀) = 1

Amount remaining (N) = 0.68

Percentage remaining = N/N₀ × 100

Percentage remaining = 0.68/1 × 100

Therefore, the percentage amount of radium-226 that remains after 900 years is 68%

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Answer:

Neutral

Explanation:

pKa of acid = -log Ka

                   = -log (1.8 x 10^-5)

                  = 4.74

pKb of base = -log Kb

                      = 4.74

pKa of acid = pKb of base

salt pH formula   : pH = 7 + 1/2 [pKa -pKb ]

here pKa = pKb

so pH = 7

the salt it is CH3COONH4 exactly neutral solution .

If a salt is formed by combining NH₃ (Kb=1.8×10⁻⁵) and CH₃COOH (Ka=1.8×10⁻⁵), an aqueous solution of this salt would be neutral.

pH of any solution tells about the acidity or basicity or neutral nature of the solution.

pH of any solution is directly proportional to the acid dissociation constant value (Ka) and base dissociation constant (Kb). In the question it is given that,

Value of Kb for NH₃ = 1.8×10⁻⁵

Value of Ka for CH₃COOH =  1.8×10⁻⁵

Ka & Kb values for the base and acid is same means it dissociates with same extent. So the aqueous solution of this acid and base is a neutral in nature as they have same number of acid and base ions in it.

Hence resultant solution will be a neutral solution .

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Answer:

pH of the solution is 2.0

Explanation:

The lactic acid is a weak acid that is in equilibrium with water as follows:

Lactic acid + H2O ⇄ Lactate + H₃O⁺

And Ka for lactic acid: 1.38x10⁻⁴

Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]

Initial concentration of lactic acid is (MW: 112.06g/mol):

80g * (1mol / 112.06g) / 1L = 0.714M

The equilibrium concentration of the species in the equilibrium are:

[Lactate] = X

[H₃O⁺] = X

[Lactic acid] = 0.714-X

Replacing in Ka expression:

1.38x10⁻⁴ = [X] [X] / [0.714-X]

9.8532x10⁻⁵ -  1.38x10⁻⁴X = X²

9.8532x10⁻⁵ -  1.38x10⁻⁴X - X² = 0

Solving for X:

X = -1.0x10⁻². False solution, there is no negative concentrations

X = 9.86x10⁻³M. Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 9.86x10⁻³M

and pH = -log [H₃O⁺] = -log 9.86x10⁻³M

pH = 2.0

Answer:

Explanation:

The equation for the decay is given as;

¹⁴₆C  --> X + ⁰₋₁e

For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;

Mass number;

14 = x + 0

x = 14

Atomic Number;

6 = x + (-1)

x = 6 + 1 =7

¹⁴₆C  --> ¹⁴₇N + ⁰₋₁e

The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N

Answer:

THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C

Explanation:

1604 J of heat is added to 48.9 g of hexane

To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.

Since 1604 J of heat = 48.9 g of hexane

Molar mass of hexane = 86 g/mol = 1 mole

then;

1604 J = 48.9 g

x = 86 g

x = 1604 * 86 / 48.9

x = 4205.4 J

Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.

In other words,

heat = molar heat capacity * temperature change

molar heat capacity = heat/ temperature change

Molar heat capacity = 4205.4 J / 14.5 C

Molar heat capacity = 290.027 J/C

The molar heat capacity of hexane is 290.027 J/ C

Answer:

30.12 mL.

Explanation:

We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:

Phosphoric acid H3PO4 will dissociate in water as follow:

H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)

From the balanced equation above,

1 mole of H3PO4 produces 3 moles of H+.

Therefore, XM H3PO4 will produce 8.70 M H+ i.e

XM H3PO4 = 8.70/3

XM H3PO4 = 2.9 M.

Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.

Next, we shall write the balanced equation for the reaction. This is illustrated below:

2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, H3PO4 (nA) = 2

Mole ratio of the base, Ba(OH)2 (nB) = 3

Data obtained from the question include the following:

Molarity of base, Ba(OH)2 (Mb) = 6.50 M

Volume of base, Ba(OH)2 (Vb) =.?

Molarity of acid, H3PO4 (Ma) = 2.9 M

Volume of acid, H3PO4 (Va) = 45 mL

The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.9 x 45 / 6.5 x Vb = 2/3

Cross multiply

2 x 6.5 x Vb = 2.9 x 45 x 3

Divide both side by 2 x 6.5

Vb = (2.9 x 45 x 3) /(2 x 6.5)

Vb = 30.12 mL

Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL

The words given are not clear, so the clear question is as follows:

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:

A. β-1,4- and α-1,6-glycosidic

B. α-1,4-glycosidic

C. α-1,4-galactose

D. an unbranched glucose

E. a branched fructose

F. α-1,6-glycosidic

Amylose is ......... polymer of ....... units joined by ........ bonds.

Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answer:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Explanation:

Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.

Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.

Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.

Hence, the correct answers in the sequential order are:

Amylose:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

Amylopectin:

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Answer:

- Addition of Ba(OH)2: favors the formation of a precipitate.

- Undergo a chemical reaction forming soluble species.

- Addition of CuSO4 : favors the formation of a precipitate.

Explanation:

Hello,

In this case, since the dissociation reaction of barium sulfate is:

[tex]BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)[/tex]

We must analyze the effect of the common ion:

- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

- By adding sodium nitrate, the following reaction will undergo:

[tex]BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)[/tex]

So the precipitate will turn into other soluble species.

- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

All of this is supported by the Le Chatelier's principle.

Best regards.

The total kinetic energy of a body is known as Thermal energy. Option A

Thermal energy is the direct sum of all the available random kinetic energies of molecules.

Also note that thermal energy is directly proportional to temperature in Kelvin.

Thus, the total kinetic energy of a body is known as Thermal energy. Option A

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Answer:

A.) Thermal energy

Explanation:

I got it correct on founders edtell

Answer:

Electrical current drives nonspontaneous reactions in electrolytic cells.

Explanation:

Electrochemical cells are cells that produce electrical energy from chemical energy.

There are two types of electrochemical cells; voltaic cells and electrolytic cells.

A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.

Answer:

electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.

Explanation:

Answer via Educere/ Founder's Education

Answer:

-170.65

188.8+ 256.8-205.8-(2x205.2)

-170.65 is the entropy change.

Entropy trade is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic gadget. It is associated with the conversion of heat or enthalpy completed in work. A thermodynamic device that has extra randomness means it has high entropy.

Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each extended by using their suitable stoichiometric coefficients, to reap ΔS° for the reaction.

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Explanation:

C5H10O2 + NaOH = C2H5COONa + C2H5OH

your result are : sodium propanoate and ethanol

A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.

Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.

The base hydrolysis of methyl butanoate is shown as,

C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH

Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).

Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.

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Answer: D - 126.4g

Explanation:  

% Yield = Actual Yield/Theoretical Yield

38% = Actual Yield/332.5

38/100 = Actual Yield/332.5

(.38)(332.5) = 126.35 g = 126.4 g Actual Yield

Answer:

is D. the correct answer

Explanation:

I'm not sure if it is. Please let me know if I'm mistaking.

Answer:

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]

[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Answer:

4.84g of FeO is the theoretical yield

Explanation:

The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:

2Fe(s) + O₂(g) → 2FeO

Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.

To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:

Moles Fe (Molar mass: 55.845g/mol)

Using the molar mass of the compound we can convert grams to moles, thus:

3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe

Moles and mass of FeO

As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.

Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:

0.0673 moles FeO × (71.844g / mol) =

Answer:

To calculate the [OH-] in the solution we must first find the pOH

That's

pH + pOH = 14

pOH = 14 - pH

First to find the pH we use the formula

From the question

[H3O+]= 2.6 × 10^-8 M

pH = - log 2.6 × 10^-8

pH = 7.6

pH = 8

So we pOH is

pOH = 14 - 8 = 6

To find the [OH-] we use the formula

6 = - log [OH-]

Find antilog of both sides

The solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7

Hope this helps you

Answer:

a. Sodium cyclopentanecarboxylate

b. No reaction

Explanation:

In this case, in the cyclopentanecarboxylic acid we have a carboxylic acid functional group. Therefore we have an "acid". The acids by definition have the ability to produce hydronium ions ([tex]H^+[/tex]).

With this in mind, for molecule a. we will have an acid-base reaction, because NaOH is a base. When we put together an acid and a base we will have as products a salt and water. In this case, the products are  Sodium cyclopentanecarboxylate (the salt) and water.

For the second molecule, we have the hydronium ion  ([tex]H^+[/tex]). This ion can not react with an acid. Because, the acid will produce the hydronium ion also, so a reaction between these compounds is not possible.

See figure 1

I hope it helps!

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

[tex]MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\[/tex]

Then, we add the half reactions:

[tex]2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0[/tex]

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

Answer:

35 hrs

Explanation:

half life of the substance [tex]t_{1/2 }[/tex] = 8.1 hr

initial amount [tex]N_{0}[/tex] = 75 g

The final amount [tex]N[/tex] = 3.9 g

The time elapsed [tex]t[/tex] = ?

we use the relationship

[tex]N[/tex] = [tex]N_{0}[/tex] [tex](\frac{1}{2} )^{\frac{t}{t_{1/2} } }[/tex]

substituting values, we have

3.9 = 75 x [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

0.052 = [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

take the log of both side

log 0.052 = log  [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

log 0.052 = [tex]\frac{t}{8.1}[/tex] log 1/2

-1.284 =  [tex]\frac{t}{8.1}[/tex]  x -0.301

1.284 = 0.301t/8.1 =

1.284 = 0.0372t

t = 1.284/0.037 = 34.5 ≅ 35 hrs

Answer:

The reaction given is:

In (g) + H₂ (g) ⇔ InH₂ (g), the Kp is 1.48 at 973 K.

The partial pressures measured in the reaction vessel is Partial pressure of In is 0.0540 atm, partial pressure of H₂ is 0.0250 atm, and the partial pressure of InH₂ is 0.0780 atm. By using the table given in the attachment below, the value of PInH₂ is (0.078-x), PIn is (0.054 + x), and the value of PH2 is (0.025 + x).

Kp = PInH₂/PIn × PH₂ = (0.078 - x) / (0.054 +x) (0.025 + x)

1.48 = (0.078 - x) / (0.054 +x) (0.025 + x)

x = 0.06689

Now the partial pressures of In, H₂ and InH₂ will be,

PH₂ = 0.025 + x = 0.025 + 0.0668 = 0.0918 atm

PIn = 0.054 + 0.0668 = 0.1208 atm

PInH₂ = 0.078 - 0.0668 = 0.0112 atm

Now the Qp or the reaction quotient will be,

Qp = (0.078) / (0.054) (0.025) = 57.78.

Answer:

-434.14 kJ

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Calculate the standard free energy change (ΔG°r) for the reaction

We will use the following expression.

ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))

ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)

ΔG°r = -959.42 kJ

Step 3: Calculate the standard free energy change for 1.81 moles of NH₃

959.42 kJ are released per 4 moles of NH₃.

[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]

Answer:

one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.

Explanation:

H-N=N-H

Answer: Km = 10μM

Explanation: Michaelis-Menten constant (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: Turnover number (Kcat), which is the number of time an enzyme site converts substrate into product per unit time.

Enzyme veolcity is calculated as:

[tex]V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}[/tex]

where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;

To calculate Km:

[tex]V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate][/tex]

[tex]K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}[/tex]

[tex]K_{m} = \frac{0.02*600*40-9.6*40}{9.6}[/tex]

Km = 10μM

The Michaelis-Menten for the substrate SAD is 10μM.

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

[tex]left\ over=0g[/tex]

Regards.

Answer:

[tex]M_{base}=0.709M[/tex]

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

[tex]n_{acid}=n_{base}[/tex]

That in terms of molarities and volumes is:

[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]

Thus, solving the molarity of the base (KOH), we obtain:

[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]

Regards.

Answer:

The sun

Explanation:

The sun is the primary source of energy in most living communities. The producers or the green plants that prepare their own food by the use of sunlight and other natural resources. Carbon dioxide, water, and other minerals are used by the plants to make their food in the presence of chlorophyll. Plants are then consumed by the consumers. This chain helps in forming the food chain and the food web.

Answer:

Boiling

Explanation:

When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.

When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.

Answer:

[tex]T_b=-88.48\°C[/tex]

Explanation:

Hello,

In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:

[tex]\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}[/tex]

Solving for the boiling point of nitrous oxide, we obtain:

[tex]T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K[/tex]

Which in degree Celsius is also:

[tex]Tb=184.67-273.15\\\\T_b=-88.48\°C[/tex]

Best regards.

Answer:

(D) The reaction cannot take place in a single elementary step

Explanation:

Statements are:

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I].

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution.

(D) The reaction cannot take place in a single elementary step.

The rate of the reaction is:

rate = k[CH3I][N3–].

That means rate depends of concentration of CH₃I as much as N₃⁻ concentration

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. FALSE. The reaction rate depends of N₃⁻ as much as CH₃I

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. FALSE. Reaction is second-order. Half-life is 1/K[A]₀. If initial concentration is 0.1M, to a concentration of 0.005M it takes:

1/K*0.1. If initial concentration is 0.005M it takes 1/K*0.005. That means it takes half to decrease from 0.005M to 0.0025 as it does for it to decrease from 0.01M to 0.005M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution. FALSE. Reaction rate is independent of I⁻

(D) The reaction cannot take place in a single elementary step. TRUE. As this reaction is a single-replacement reaction implies the formation  of 1 C-N bond. But also the rupture of the C-I bond is impossible to explain this kind of reaction in a single elementary step.