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2 years ago

Do organic compounds ionize? yes or no

2 answers:

5 0

Surface ionization of organic compounds in a weak external electric field involving the formation of many-atomic positive ions is considered (including specific features of the phenomenon, its major characteristics, experimental techniques used and possible applications).

Stella [2.4K]2 years ago

3 0

Answer:

Yes but it's in a weak force

Explaination:

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The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b

kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$ :

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

6 0

3 years ago

What is the coefficient of mno2 when you balance the equation for this redox reaction? kcn(aq) + kmno4(aq) + h2o(l) → kcno(aq) +

Tasya [4]

Kcn(aq)+kmno4(aq)+h2o(l)----->2mno2(s)+koh(aq)

7 0

3 years ago

Read 2 more answers

Container A (with volume 1.23 L) contains a gas under 3.24 atm of pressure. Container B (with volume

kolbaska11 [484]

Now that a lot of pressure

8 0

3 years ago

Which of the following 0.820 M solutions would have the greatest colligative effect?

eimsori [14]

Answer:

K3PO4

Explanation:

Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;

SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)

C4H11N (not ionic in nature hence it can not dissociate into ions)

K3PO4-------> 3K^+ + PO4^3- (4 particles)

Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)

Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.

8 0

3 years ago

Do organic compounds ionize? yes or no​

Do organic compounds ionize? yes or no