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Sonja [21]
3 years ago
7

The valency of both Oxygen and magnesium is 2 give reason ​

Chemistry
1 answer:
lidiya [134]3 years ago
7 0

Answer:

<em>the <u>valency of an element</u> is its combining capacity that is the number of electrons it requires to lose, gain or share in order to become neutral.</em>

[ An element can become neutral if it completes it's octet. That is if an element has 8 electrons in it'd outermost shell then it is considered neutral ]

  • The valence of Magnesium is 2 because it requires to lose 2 electrons to become neutral.

  • whereas, the valence of Oxygen is 2 because it needs to gain 2 electrons to become stable.

Hence they both have the same valence.

One may say that oxygen's valence is -2 while that of Magnesium is + 2. It's meaning is still the same but "-" sign indicates that oxygen will be gaining electrons in the process of becoming stable.

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B) is the correct answer

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What were the defects in Ruther Ford Atomic model?
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the RutherFord atomic model has the limitations in explaining the stability of the atom and the stability of the electron.

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9. How many significant figures are there in the number 0.0001372?
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3 years ago
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0
4 years ago
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