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The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answers

Answer 1

Answer:

Explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s

Answer 2

Answer:

Answer:

(a)

$0.0342M$

(b)

$t_(1/2)=17.36s\nt_(1/2)=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$(1)/([NOBr])=kt +(1)/([NOBr]_0)\n(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\n$

$[NOBr]=(1)/(29.2/M)=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_(1/2)=(1)/(k[NOBr]_0)$

Therefore, for the given initial concentrations one obtains:

$t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\nt_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s$

Best regards.

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What is the polarity of a Na-Br bond?nonpolar covalent

definitely polar covalent

likely ionic

slightly polar covalent

slightly ionic

Answers

Answer:

likely ionic

Explanation:

The bond between , Sodium and Bromine , in sodium bromide salt , is ionic in nature .

Since ,

The difference in the electronegativity of the two atom is high .

According to Pauling scale ,

The Electronegativity of Bromine = 2.96

The Electronegativity of sodium = 0.93

Hence ,

The Difference in electronegativity = ( 2.96 - 0.93 ) = 2.03

Hence ,

The electronegativity difference value is more than 1.8 ,

Therefore,

The bond is ionic in nature .

A cool, yellow-orange flame is used to heat the crucible. Would this affect the mass of the crucible? If so, how?

Answers

Answer:

yes

Explanation:

Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.

Final answer:

Using a cool, yellow-orange flame to heat the crucible does not directly affect its mass, but can lead to the burning off or decomposition of any impurities or residues present.

Explanation:

When a cool, yellow-orange flame is used to heat the crucible, it does not directly affect the mass of the crucible. The color of the flame is an indication of the temperature and the type of fuel being burned.

However, if there are impurities or residues in the crucible, the heat from the flame can cause them to burn off or decompose, which may slightly affect the mass of the crucible.

Learn more about Effect of flame color on crucible mass here:

brainly.com/question/31815558

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Is A and C correct?

I'm a little unsure of this question, but is it right?