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mariarad [96]
3 years ago
6

What is the conjugate acid for the base, NH4?​

Chemistry
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

NH4+ is the conjugate acid of the base NH3.

Explanation:

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Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
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Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

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The state of matter with the slowest moving atoms is _
devlian [24]
The state of matter with the slowest moving atoms is solid!
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(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the
djyliett [7]

Answer:

Explanation:

1. The amount of CaCO3 must be so small that  

P

CO

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P

CO

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3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)  

K

c

=

[

CH

3

OH

]

[

H

2

]

2

[

CO

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)  

K

c

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[

CO

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[

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2

]

[

H

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; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

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