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3 years ago

What determines the most likely oxidation state of an element?

Chemistry

2 answers:

Salsk061 [2.6K]3 years ago

8 0

C. The number of electrons the element needs to lose or gain to have a full valence shell

jekas [21]3 years ago

7 0

Answer: The correct answer is Option C.

Explanation:

Oxidation state is defined as the number which is given to an atom when it looses or gains electron. It is written as a superscript.

When an atom looses electron, it will attain a positive oxidation state and when an atom gains electron, it will attain a negative oxidation state.

An atom looses or gains electron to attain stability and to fill their valence shell.

For Example:

Oxygen atom will gain 2 electrons and will form $O^{2-}$ ion having oxidation state of -2.
Sodium atom will loose 1 electron and form $Na^+$ ion having oxidation state of +1.

Thus, the correct answer is Option C.

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A substance that does not allow x-rays to pass through is described as being

horsena [70]

A substance that is impenetrable by x-rays is described as being radiopaque.

Radiopaque substances will not allow x-rays and or other forms of radiations to pass through them.

Instead, they absorb or block the rays and when used in radiology, they appear white or light gray on photographic films.

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2 years ago

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adelina 88 [10]

Answer:

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Explanation:

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3 years ago

What group is helium in? Group 8 or group 2?

Gemiola [76]

Answer:

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Explanation:

7 0

3 years ago

Read 2 more answers

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of $HF$ .

$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

Initial moles 0.375 0.100 0.375

At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

Now put all the given values in this expression, we get:

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

8 0

3 years ago

Sliva [168]

From the balanced equation:
4mol Fe will produce 2mol Fe2O3
Molar mass Fe = 55.847g/mol
16.7gFe = 16.7/55.847 = 0.3mol Fe
This will produce 0.15mol Fe2O3
Molar mass Fe2O3 = 159.6887 g/mol
0.15mol = 159.6887*0.15 = 23.95g Fe2O3 produced
Hope this helps

6 0

3 years ago