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marysya [2.9K]
3 years ago
15

The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

1. If the initial pressure of N2O is 5.10 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant.
Chemistry
1 answer:
fomenos3 years ago
4 0

Answer:

The total pressure after one half is 6.375 atm.

Explanation:

The initial pressure of product is increases while the pressure of reactant would decrease.

Balanced chemical equation:

2N₂O  → 2N₂ + O₂

The pressure of N₂O is 5.10 atm. The change in pressure would be,

N₂O = -2x

N₂ = +2x

O₂ = +x

The total pressure will be

P(total) = P(N₂O) + P(N₂)  + P(O₂)

P(total) = ( 5.10 - 2x) + (2x)  + (x)

P(total) = 5.10 + x

After one half life:

P(N₂O)  = 1/2(5.10) = 5.10 - 2x

x = 5.10 - 1/2(5.10) /2

x = 5.10 - 0.5 (5.10) /2

x = 5.10 - 2.55 / 2

x = 2.55 /2 = 1.275 atm

Thus the total pressure will be,

P(total) = 5.10 + x

P(total) = 5.10 + 1.275

P(total) = 6.375 atm

You might be interested in
A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask
leonid [27]

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer:  The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of Ba(CH_3COO)_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.125\times 1000}{350ml}

Molarity=0.357M

Therefore, the concentration of solution is 0.375 mol/L

8 0
3 years ago
(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

6 0
3 years ago
Select all possible answers:
Dafna11 [192]

Answer:

c and d are correct

Explanation:

In A, false because in Valence Electrons, the more the valences, the more stable an atom is.

In B, false because atoms cannot readily gain or lose valence electrons as the number of valence electrons is determined by the column they are in.

In C, true because the more the valence electrons, the more the stability of an atom.  

In D, true as electron placing is important and the reactivity of an atom is important.

So C and D are true!

7 0
3 years ago
Read 2 more answers
Please help with this
dlinn [17]

Answer:

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

Step by Step Explanation?

Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.

The formula is,

P₁ x V₁ = P₂ × V₂

Where,

P₁ is initial pressure = 1 atm

P2 is final pressure = ? (Not Known)

V₁ is initial volume = 10 L

V₂ is final volume = 15 L

Now put the values in the formula,

\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]

Therefore, the answer is 0.67 atm.

5 0
2 years ago
Read 2 more answers
What does sodium react with
Pani-rosa [81]
Sodium reacts with water to form a colorless solution of sodium hydroxide and hydrogen gas.
5 0
3 years ago
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