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GenaCL600 [577]
3 years ago
11

Why is "Al" and "o" in iconic bonding equal to "Al2O3"

Chemistry
1 answer:
Yuri [45]3 years ago
4 0
Both of these elements have a charge based on the number of valence electrons. For example: Nitrogen has 3 more electrons (negative) than protons (positive) so it has a negative 3 charge.
Aluminum has a positive 3 charge (three more protons than electrons)
Oxygen has a negative 2 charge (two more electrons than protons)
In order to put these two elements together, they have to balance each other out so that way their compound has no charge.
So for example if you are combining Nitrogen with Hydrogen, you need to balance the charges. So what you do is since Nitrogen has a negative 3 and Hydrogen has a positive 1 you need three hydrogens (H3) to balance out one Nitrogen. so the compound would be H3N.
With this compound you need to have 2 Aluminum (which makes a positive 6 charge) and three Oxygen (which makes a negative 6 charge). When you combine them there is no longer a charge.
Basically think of it as simple addition where you're just trying to make 0. So 6+(-6)=0. This makes no charge.
I hope this kinda helped but please ask questions if you need any more help :)

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Regular table salt is sodium chloride, NaCl. It is a bonding of an Na+ ion and a Cl- ion. Which of these ions is bigger?
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A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55
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<u>Answer:</u> The volume of NaOH required is 402.9 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For HCl:</u>

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol

  • <u>For sulfuric acid:</u>

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL

Hence, the volume of NaOH required is 402.9 mL

6 0
3 years ago
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