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Naddika [18.5K]
3 years ago
13

If two electrons are removed from beryllium during a reaction, which energy level becomes the valence shell?

Chemistry
1 answer:
tino4ka555 [31]3 years ago
3 0

Explanation:

After the electron configuration, the last shell of the beryllium atom has two electrons. In this case, both the valence and valence electrons of beryllium are 2. We know the details about this. The elements that have 1, 2, or 3 electrons in the last shell donate the electrons in the last shell during bond formation.

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A sample of propane(c3h8)has 3.84x10^24 H atoms.
deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

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If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}

<u>= 4.75 × 10²³ molecules of Propane</u>.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

<u>= 14.25 × 10²³ </u>

<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

<u>= 34.72 g</u>

8 0
2 years ago
Learning calculations in chemistry
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What question do u have?
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3 years ago
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