All categories

3 years ago

Determine the hybridization around the central atom for each of the following molecules.

Chemistry

1 answer:

alina1380 [7]3 years ago

3 0

Answer:

a) HCN - hybridization sp

b) C(CH₃)₄ - hybridization sp³

c) H₃O⁺ - hybridization sp³

d) - CH₃ - hybridization sp³

Explanation:

Hybridization occurs to allow an atom to make more covalent bonds than the original electronic distribution would allow or to allocate ligands in an energetically stable geometry.

Carbon can have thre hybridization states: sp³ , sp² and sp.

Oxygen usualluy has an sp³ hybridization.

In order to determine the hybridization, we need to consider the number of atoms attached to the central atom and the number of lone pairs.

The figure attached shows the species and the hybridization of their central atoms.

You might be interested in

Consider the formation of hcn by the reaction of nacn (sodium cyanide) with an acid such as h2so4 (sulfuric acid): 2nacn(s)+h2so

Artemon [7]

2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)

The molar ratio between NaCN : HCN is 2:2 or 1:1

Mass of HCN = 16.7 g

Molar mass of HCN = 1 + 12 + 14 = 27 g/mol

Molar mass of NaCN = 49 g/mol

Therefore, the mass of NaCN is

16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN

Therefore, 30.3 grams of NaCN gives the lethal dose in the room.

5 0

3 years ago

Read 2 more answers

Suppose that 25.0 mL of 0.440 M sodium chloride is added to 25.0 mL of 0.320 M silver nitrate. How many moles of silver chloride

d1i1m1o1n [39]

The number of moles of silver chloride that will precipitate is 0.008 mole

From the question,

We are to determine the number of moles of silver chloride that will precipitate

First,

We will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

NaCl + AgNO₃ → AgCl + NaNO₃

This means

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride and 1 mole of sodium nitrate

Now, we will determine the number of moles of each reactant present

For sodium chloride (NaCl)

Concentration = 0.440 M

Volume = 25.0 mL = 0.025 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of NaCl present = 0.440 × 0.025

Number of moles of NaCl present = 0.011 mole

For silver nitrate (NaNO₃)

Concentration = 0.320 M

Volume = 25.0 mL = 0.025 L

∴ Number of moles of NaNO₃ present = 0.320 × 0.025

Number of moles of NaNO₃ present = 0.008 mole

From the balanced chemical equation,

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride

Then,

0.008 mole of sodium chloride will react with the 0.008 mole of silver nitrate to produce 0.008 mole of silver chloride

∴ 0.008 mole of silver chloride will be produced

Hence, the number of moles of silver chloride that will precipitate is 0.008 mole

Learn more here: brainly.com/question/18434602

4 0

2 years ago

H-C C-H name of formula

Vanyuwa [196]

Answer:

Ethyne

Explanation:

7 0

3 years ago

How many grams of sodium phosphate are formed by the reaction of 10 g of sodium with phosphoric acid?

Norma-Jean [14]

Answer:

23.8g of sodium phosphate are formed

Explanation:

Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:

3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂

<em>3 moles of sodium produce 1 mole of sodium phosphate</em>

<em />

To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:

<em>Moles Na -Molar mass: 22.99g/mol-</em>

10g * (1mol / 22.99g) = 0.435 moles Na

<em>Moles Na₃PO₄:</em>

0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄

<em>Mass Na₃PO₄ -Molar mass: 163.94g/mol-</em>

0.145 moles Na₃PO₄ * (163.94g/mol) =

<h3>23.8g of sodium phosphate are formed</h3>

8 0

2 years ago

0.0027432 using scientific notation

alexira [117]

Answer: 2.7423 × 10^-3

Explanation:

4 0

3 years ago