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notka56 [123]
3 years ago
13

A possible mechanism for the overall reaction br2 (g) + 2no (g) → 2nobr (g) is no (g) + br2 (g) nobr2 (g) (fast) nobr2 (g) + no

(g) k2→ 2nobr (slow) the rate law for formation of nobr based on this mechanism is rate = ________. a possible mechanism for the overall reaction br2 (g) + 2no (g) 2nobr (g) is no (g) + br2 (g) nobr2 (g) (fast) nobr2 (g) + no (g) 2nobr (slow) the rate law for formation of nobr based on this mechanism is rate = ________. k1[br2]1/2 (k2k1/k-1)[no][br2]2 (k1/k-1)2[no]2 k1[no]1/2 (k2k1/k-1)[no]2[br2]
Chemistry
1 answer:
givi [52]3 years ago
7 0
The overall reaction is:
Br₂(g) + 2 NO(g) ↔ 2 NOBr(g)
rate law = k [Br₂][NO]²

The first step of the overall reaction is:
NO(g) + Br₂(g)   K₁⇄⇄K-1  NOBr₂(g)
rate law 1 = k₁ [Br₂][NO] or 
rate law 2 = k-1 [NOBr₂]

The second step of the overall reaction is:
NOBr₂(g) + NO(g)  →K₂→ 2 NOBr
rate law 3 = k₂[NOBr₂][NO]

So, rate law of overall reaction can be obtained as follows:
(rate law 1)*(rate law 3) / (rate law 2)
= [(k₁[Br₂][NO])* (K₂[NOBr₂][NO])] / k₋₁[NOBr₂]
= [k₁k₂/k₋₁][NO]²[Br₂]
So the correct answer is:
[k₁k₂/k₋₁][NO]² [Br₂]
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1. 90.0 mL of distilled water is added to a 10.0mL sample of 0.150mol/L sodium
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Answer: dilute

Explanation:

A concentrated solution which is used to prepare solutions of lower concentrations by diluting it with addition of water.

A dilute solution is one which contains lower concentration.

Using Molarity equation:

M_1 =concentration of stock solution = 0.150 mol/L

V_1 = volume of stock solution = 10.0 ml

M_2 = concentration of dilute solution = ?

V_2 = volume of dilute solution = (10.0+90.0) ml = 100.0 ml

0.150\times 10.0=M_2\times 100.0

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As the concentration is less than the original concentration, the solution is termed as dilute.

7 0
2 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

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