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balu736 [363]
3 years ago
8

The hypochlorite ion (OCl-) is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the acti

ve ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-, for example) and forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 x 10-8). Calculate the pH of a 0.100 M aqueous solution ofhypochlorous acid.
Chemistry
1 answer:
Yanka [14]3 years ago
6 0

Answer:

In Addition To Its Oxidizing Abilities, The Hypochlorite Ion Has A Relatively High Affinity For ... It is also the active ingredient that forms when swimming pool water is treated with chlorine.

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The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air
IceJOKER [234]

Answer:

1.75\cdot 10^{-4} M

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

S = k_H p^o

Where k_H is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

S_1 = 2.67\cdot 10^{-4} M

Also, at sea level, we have an atmospheric pressure of:

p = 1.00 atm

Given mole fraction:

\chi_{O_2} = 0.209

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

p^o = \chi_{O_2} p

Then the equation becomes:

S_1 = k_H \chi_{O_2} p

Solve for k_H:

k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

p = 0.657 atm

Apply Henry's law using the constant we found:

S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M

8 0
3 years ago
Xbox or ps4? haha have a. Good ond
bekas [8.4K]

Answer:

I play none but If I did I would choose Xbox

btw, thank you

7 0
3 years ago
Preserving food by freezing controls reaction rates by
kykrilka [37]

Answer:

D) both a and c are correct

Explanation:

The reaction rate is a measure of the speed of a chemical reaction. The factors that affects the rate of a chemical reaction are itemised below:

  • Nature of the reactants
  • Concentration of the reactants or pressure(if gaseous)
  • Temperature
  • Presence of catalyst
  • Sunlight

Our concern here is temperature. Temperature affects a reaction considerably. Average kinetic energy is directly proportional to the temperature of the reacting particles. When the temperature of a reacting system is increase, the frequency of ordinary and effective collisions per unit time increases. A decrease in temperature implies that the number of collisions also decreases.

3 0
3 years ago
You place 12.0 milliliters of water in a graduated cylinder. Then you add 15.0 grams of metal to the water and the new water lev
Marrrta [24]

Answer:

<h3>The answer is 7.85 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

volume = final volume of water - initial volume of water

volume = 13.91 - 12 = 1.91 mL

We have

density =  \frac{15}{1.91}  \\  = 7.853403141...

We have the final answer as

<h3>7.85 g/mL</h3>

Hope this helps you

8 0
3 years ago
Read 2 more answers
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
3 years ago
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