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3 years ago

Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.

Chemistry

1 answer:

mars1129 [50]3 years ago

7 0

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

Ka= 0.2(0.001²)/(1-0.001)

Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

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Problem PageQuestion The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium

Shalnov [3]

Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The reaction that takes place is:

2NaN₃(s) → 2Na(s) + 3N₂(g)

2. We use PV=nRT to calculate the moles of N₂ that were produced.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 K

n = 0.334 mol

Now we convert N₂ moles to NaN₃ moles:

0.334 mol N₂ * $\frac{2molNaN_{3}}{3molN_2}$ = 0.223 mol NaN₃

Finally we convert NaN₃ moles to grams, using its molar mass:

0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃

6 0

3 years ago

How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.

Slav-nsk [51]

Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

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3 years ago

What's the correct answer

svp [43]

The correct answer I believe is sunlight

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3 years ago

Which of the following might a chemist choose to study?

Serjik [45]

Answer:

B. Better ways to recycle plastics

Explanation:

A chemist might choose to study better ways to recycle plastics because there may be certain types of chemicals that can disintegrate the plastic..

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3 years ago

What can be explained by the Arrhenius theory?

vekshin1

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