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KiRa [710]
3 years ago
8

#12 PLEASE HELP IM TAKING A TIMED TEST

Chemistry
1 answer:
Goshia [24]3 years ago
3 0

Answer:

Option B

Explanation:

For the problem in question 12, option B is the correct choice. It rightly expresses how electrons will fill the third energy level of aluminum.

Aluminum is an element with 13 electrons. It has 3 energy levels.

 The shell notation is given as 2 8 3

 Sublevel notation  = 1s² 2s² 2p⁶ 3s² 3p¹

 The third energy level is given as 3s² 3p¹

  We can represent it as:  ⇅     ↑

                                            3s²   3p¹

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If the pH is 10, what is the concentration of hydroxide ion?
AleksandrR [38]

Answer : The correct option is, 10^{-4}M

Explanation : Given,

pH = 10

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10=4

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

4=-\log [OH^-]

[OH^-]=1.0\times 10^{-4}M

Therefore, the OH^- concentration is, 1.0\times 10^{-4}M

7 0
3 years ago
When 0.100 mol of carbon is burned in a closed vessel with8.00
antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

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Molar mass of CO = 28 g/mole

First we have to calculate the moles of O_2.

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C+O_2\rightarrow 2CO

From the balanced reaction we conclude that

As, 2 mole of C react with 1 mole of O_2

So, 0.1 moles of C react with \frac{0.1}{2}=0.05 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and C is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO

From the reaction, we conclude that

As, 2 mole of C react to give 2 mole of CO

So, 0.1 moles of C react to give 0.1 moles of CO

Now we have to calculate the mass of CO

\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO

\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g

Therefore, the mass of carbon monoxide form can be 2.8 grams.

5 0
4 years ago
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3 years ago
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Google

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