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Maru [420]
3 years ago
10

What occurs in a half-cell? oxidation or reduction, but not both reduction only oxidation only?

Chemistry
1 answer:
lesya692 [45]3 years ago
5 0
Answer: In a half-cell is where occurs the either oxidation (loss of electrons) or reduction (gain of electrons) but not both.

Explanation:

1) In an electrolitic cell you can identify two half-cells.

2) In one half-cell occurs the oxidation

3) In the other half-cell occurs the reduction.

4) The half-cell where oxidation occurs is the anode.

In the anode the spieces increase its oxidation number, which is what oxidation means. Increasing the oxidation number means that the spieces is losing electrons.


5) The half-cell where reduction occurs is the cathode.

Reduction is the reduction of the oxidation number and this occurs due to the gain of electrons.


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What is the distance between (–6, 2) and (8, 10) on a coordinate grid? mc017-1.jpg mc017-2.jpg mc017-3.jpg mc017-4.jpg
Svetach [21]
The distance between (–6, 2) and (8, 10) on a coordinate grid is 8.246
5 0
3 years ago
Read 2 more answers
What percent of I-125 has decayed if there are 37.5g of the original sample left?
garri49 [273]

Answer:

[(x-37.5)/x]*100%

62.5%  (assuming the original sample weighs 100.0g)

Explanation:

Let's say that the original sample is x

Mass of I-125 which has decayed: x-37.5

Percentage of decayed mass: [(x-37.5)/x]*100%

Please recheck, for this may not be the correct answer

6 0
3 years ago
What is the pH of a solution with a 4.60 × 10−4 M hydroxide ion concentration?
guapka [62]

Answer:

pH

=

10.66

Explanation:

For pure water at  

25

∘

C

, the concentration of hydronium ions,  

H

3

O

+

, is equal to the concentration of hydroxide ions,  

OH

−

.

More specifically, water undergoes a self-ionization reaction that results in the formation of equal concentrations of hydronium and hydroxide anions.

2

H

2

O

(l]

⇌

H

3

O

+

(aq]

+

OH

−

(aq]

At room temperature, the self-ionization constant of water is equal to

K

W

=

[

H

3

O

+

]

⋅

[

OH

−

]

=

10

−

14

This means that neutral water at this temperature will have

[

H

3

O

+

]

=

[

OH

−

]

=

10

−

7

M

As you know, pH and pOH are defined as

pH

=

−

log

(

[

H

3

O

+

]

)

pOH

=

−

log

(

[

OH

−

]

)

and have the following relationship

pH

+

pOH

=

14

In your case, the concentration of hydroxide ions is bigger than  

10

−

7

M

, which tells you that you're dealing with a basic solution and that you can expect the pH of the water to be higher than  

7

.

A pH equal to  

7

is characteristic of a neutral aqueous solution at room temperature.

So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first

pOH

=

−

log

(

4.62

⋅

10

−

4

)

=

3.34

This means that the solution's pH will be

pH

=

14

−

pOH

pH

=

14

−

3.34

=

10.66

Indeed, the pH is higher than  

7

, which confirms that you're dealing with a basic solution.

4 0
4 years ago
Read 2 more answers
You decide to clean the bathroom. You notice that the shower is covered in a strange green slime. You try
musickatia [10]

Answer:

Use either a vinegar, Borax, or bleach solution in a spray bottle to tackle the mold. Simply spray the solution on showers, baths, basins, tiles, grout, or caulking. Then use either a cleaning cloth or a toothbrush to remove the slime

7 0
3 years ago
A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point
Ksivusya [100]

Answer:

P_{T} = 2.94 atm

Explanation:

The total pressure (P_{T}) in the container is given by:

P_{T} = P_{O_{2}} + P_{He}

The pressure of the oxygen (P_{O_{2}}) and the pressure of the helium (P_{He}) can be calculated using the ideal gas law:

PV = nRT

<u>Where</u>:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

n_{O_{2}} = \frac{m}{M}

Where m is the mass of the gas and M is the molar mass

n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles  

And the number of moles of helium is:

n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles

Now, we can find the pressure of the oxygen and the pressure of the helium:

P_{O_{2}} = \frac{nRT}{V} = \frac{1.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 0.98 atm

P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm

Finally, the total pressure in the container is:

P_{T} = P_{O_{2}} + P_{He} = 0.98 atm + 1.96 atm = 2.94 atm

Therefore, the total pressure in the container is 2.94 atm.

I hope it helps you!

6 0
4 years ago
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