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astraxan [27]
3 years ago
13

A molecule with the formula ax3 uses __________ to form its bonds. multiple choice sp2 hybrid orbitals sp hybrid orbitals sp3d2

hybrid orbitals sp3d hybrid orbitals sp3 hybrid orbitals
Chemistry
2 answers:
Ivanshal [37]3 years ago
8 0
A molecule with formula AX3E uses SP3 to form its bonds.
Where A is the metal
X is any halogen which is bonded to it.
E is the lone pair
∴ A requires four electron pairs and tetrahedral.
Where tetrahedral symmetry is SP3 hybridized
GenaCL600 [577]3 years ago
4 0

A molecule with formula AX3E uses SP3 to form its bonds.

Where A is the metal

X is any halogen which is bonded to it.

E is the lone pair

A requires four electron pairs and tetrahedral.

Where tetrahedral symmetry is SP3 hybridized

Read more on Brainly.com - brainly.com/question/9810726#readmore

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Classify the following reaction.
Klio2033 [76]

Answer:

Explanation:

combustion  reacts with oxygen

7 0
3 years ago
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.

The sample has 3.84 × 10²⁴ H atoms.

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
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Answer:

In diagram you have to show the electrons(dots) near to the more electronegative element by this you can show that this is an ionic bond.

Explanation:

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Calculate the number of ATOMS in 1.0 mole of O2.
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There is 6.02*10^23 molecule per mole. And there is 2 atoms per oxygen molecule. So the answer is 1.204*10^24 atoms in 1.0 mole of O2.
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