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Katarina [22]
3 years ago
13

Is it possible for a diprotic acid to have Ka1?

Chemistry
1 answer:
Masja [62]3 years ago
4 0

Answer:

Yes, it is possible.

Explanation:

A diprotic acid is an acid that can release two protons. That's why it is called diprotic.

Monoprotic → Release one proton, for example Formic acid HCOOH

Triprotic → Releases three protons, for example H₃PO₄

Polyprotic → Release many protons, for example EDTA

it is a weak acid.

In the first equilibrum, it release proton, and the second is released in the second equilibrium. So the first equilibrium will have a Ka1

H₂A  +  H₂O  ⇄  H₃O⁺  +  HA⁻                Ka₁

HA⁻  +  H₂O  ⇄  H₃O⁺  +  A⁻²               Ka₂

The HA⁻ will work as an amphoterous because, it can be a base or an acid, according to this:

HA⁻   +  H₂O  ⇄  H₃O⁺   + A⁻²      Ka₂

HA⁻   +  H₂O  ⇄  OH⁻   + H₂A       Kb₂

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Suppose the decomposition of dinitrogen monoxide proceeds by the following mechanism:
photoshop1234 [79]

Answer:

Part A: 2N₂O(g) ⇄ 2N₂(g) + O₂(g)

Part B: -r = K*[N₂O]²

Part C: K= k1*k2

Explanation:

Part A

To do the balance chemical question for the overall chemical reaction, we must sum the reaction of the steps, eliminating the intermediaries, which are the compounds that have the same amount both at reactants and products (bolded).

N₂O(g) ⇄ N₂(g) + O(g)

N₂O(g) + O(g) ⇄ N₂(g) + O₂(g)

---------------------------------------------

2N₂O(g) + O(g) ⇄ 2N₂(g) + O(g) + O₂(g)

2N₂O(g) ⇄ 2N₂(g) + O₂(g)

Part B

The velocity of the reaction (r) can be calculated based on the reactants or based on the products. Let's do it based on the disappearing of the reactant. Because it is disappearing, the variation at its concentration must be negative, so the rate will be negative.

Let's suppose its an elementary reaction, so, the concentration of the reactant must be elevated by its coefficient. And let's call the overall rate constant as K:

-r = K*[N₂O]²

Part C

Because the steps were summed, and the reactions were not multiplied by a constant or inverted, the constant K is just the multiplication of the constants of the steps:

K= k1*k2

8 0
3 years ago
Which of these structures are found in both plant and animal cells?
dlinn [17]

Answer:

the cell wall for sure not to sure about the others

6 0
3 years ago
Have you ever tried to mix sugar with a cold drink like tea or lemonade? What happened? Explain why.
Vinvika [58]

Answer:

yes

Explanation:

it wont dissolve because when water is cold its molecules dont separate

5 0
3 years ago
Which is not a functional group<br><br><br>A. Carboxyl <br>B. Halide<br>C. Hydroxyl<br>D. Protein
xeze [42]
I believe it’s B. Halide
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3 years ago
A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t
leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = P_1=1.0 atm

Initial temperature of the oxygen gas = T_1=20^oC =293.15 K

Final pressure of the oxygen gas = P_2=?

Final temperature of the oxygen gas = T_2=100^oC =373.15 K

Using Gay Lussac's law:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

P_1V_1=nRT_1 (ideal gas equation)

n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

6 0
3 years ago
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