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1 year ago

How many of the following molecules have sp2 hybridization on the central atom? hcn so2 ocl2 xecl2.

Chemistry

1 answer:

GrogVix [38]1 year ago

7 0

Only one molecule have sp2 hybridization on central atom and that is SO₂.

XeCl₂ have sp3d hybridization.

OCl₂ have sp3 hybridization.

HCN have sp hybridization.

Hybridization is defined as the concept of mixing two atomic orbitals to give rise to a new type of hybridized orbitals.

Hybridization intermixing usually results in the formation of hybrid orbitals having entirely different energies, shapes, etc.

Learn more about hybridization here:- brainly.com/question/22765530

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a_sh-v [17]

The correct answer is actually

D. requires a lot of energy to become hot.

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3 years ago

Ignoring sign which transition is associated with the greatest energy change?

scoray [572]

Well u said ignoring not me

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3 years ago

Need help fast!!

stepan [7]

Answer:

The answer is "5.18 ".

Explanation:

It's a question of chemistry. Therefore, the following solution is provided:

We will first determine the solution's pOH. The following can possible:

The concentration of Hydroxide ion $[OH^{-}] = 1.5\times 10^{-9}\ M$

$pOH =?\\\\pOH = -\log [OH^{-}]\\\\pOH = -\log 1.5\times 10^{-9}\\\\pOH = 8.82\\\\$

Furthermore, the pH of the solution shall be established. It was provided as follows:

$pOH = 8.82\\\\pH =?\\\\pH + pOH = 14\\\\pH + 8.82 = 14\\\\$

When collecting all the like terms:

$pH = 14 -8.82\\\\pH = 5.18$

Therefore, the solution of pH is 5.18.

8 0

3 years ago

How many oxygen atoms in 3.161*10^21 molecules of CO2?

vfiekz [6]

Answer:

Explanation:

You need to know one piece of information for these problems. That is as follows:

There are 6.02 x 10^23 atoms in a mole of atoms or 6.02 x 10^23 molecules in a mole of molecules (actually there are 6.02 x 10^23 in a mole of anything).

So in 3.161 x 10^21 molecules of CO2 there are 3.161 molecules x (1 mole CO2/6.02 x 10^23 molecules CO2)= ?? moles CO2. Then multiply that by 2 to find the moles of O in CO2.

The others are done th same way.

3 0

4 years ago

Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.

ivanzaharov [21]

Answer:

a) 4.73

b) 4.78

c) 4.66 (further addition) or 4.60 (starting from the original buffer solution)

Explanation:

Step 1: Data given

volume of the buffer = 1.00 L

Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

pKa CH3COOH = 4.75

Step 2: pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log (0.97/1.02)

pH = 4.73

(b) pH after addition of 0.065 mol NaOH

Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = 4.75 + log(1.035/0.955)

pH = 4.78

c) pH after further addition of 0.144 mol HCl

The reverse will happen after the addition of HCl:

[HA] = 0.955 + 0.144 = 1.099

[A-] = 1.035 - 0.144 = 0.891

pH = 4.75 + log(0.891/1.099)

pH = 4.66

If we add 0.144 mol of HCl to the original buffer we will get:

[HA] = 1.02 + 0.144 = 1.164

[A-] = 0.97 - 0.144 = 0.826

pH = 4.75 + log(0.826/1.164)

pH = 4.60

3 0

3 years ago