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3 years ago

When might a plot of ln[a] vs time not yield a straight line?

1 answer:

6 0

Considering a reaction:
A → B
The rate equation may be described as:
r = -k[A]ⁿ
Taking the natural log,
ln(r) = -nln([A]) + ln(k)
Therefore, the only time the graph of ln[A] vs time will be a straight line is when the order of the reaction is 0, meaning the reaction is independent of reactant concentration.

You might be interested in

draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum

loris [4]

Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

5 0

3 years ago

If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K and then I raise the pressu

mixas84 [53]

Answer:

The answer to your question is V2 = 29.6 l

Explanation:

Data

Pressure 1 = P1 = 12 atm

Volume 1 = V1 = 23 l

Temperature 1 = T1 = 200 °K

Pressure 2 = 14 atm

Volume 2 = V2 = =

Temperature 2 = T2 = 300°K

Process

1.- To solve this problem use the Combine gas law.

P1V1/T1 = P2V2/T2

-Solve for V2

V2 = P1V1T2 / T1P2

2.- Substitution

V2 = (12)(23)(300) / (200)(14)

3.- Simplification

V2 = 82800 / 2800

4.- Result

V2 = 29.6 l

5 0

3 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

3 years ago

What element in magma is most abundant

pishuonlain [190]

Oxygen :) hope it helped

6 0

3 years ago

Read 2 more answers

In the best Lewis structure for the fulminate ion, CNO–, what is the formal charge on the central nitrogen atom?

Andrews [41]

Answer : The formal charge on the central nitrogen atom is, (+1)

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of $CNO^-$ .