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Grace [21]
2 years ago
6

Give the hybridization for the c in hcch. group of answer choices

Chemistry
1 answer:
Tamiku [17]2 years ago
5 0

The hybridization for C in acetylene, HCCH, or C₂H₂ is 'sp'.

Discussion:

There are three different forms of hybridization -

  1. sp- The first occurs when two carbon atoms are triple linked.
  2. sp₂- When two carbon atoms are double-bonded to one another, this is known as sp₂.
  3. sp₃- When a single bond joins two carbon atoms, this is known as sp₃.

In the case of acetylene(HCCH or C₂H₂):

  • The carbon atom requires additional electrons to establish four bonds with hydrogen and other carbon atoms in the synthesis of C₂H₂. One 2s₂ pair is consequently transferred to the vacant 2pz orbital. Each carbon has two sp hybrid orbitals after the 2s orbital in each atom combines with one of the 2p orbitals.
  • As a result of the atoms' symmetrical alignment in a single plane, C₂H₂ possesses a linear molecular structure. Due to their lower electronegative nature than Hydrogen atoms, all Carbon atoms are situated near the center of the Lewis structure of C₂H₂.

                                              H-C≡C-H

Therefore, it is concluded from the above discussion that the hybridization type of acetylene is 'sp'.

Learn more about hybridization here:

brainly.com/question/14140731

#SPJ4

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I need to know number 2 3 4 and 5
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Answer 2: 57.21235000000001

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7 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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