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Leni [432]
1 year ago
13

What atomic or hybrid orbitals make up the sigma bond between c2 and h in ethylene, ch2ch2?

Chemistry
1 answer:
vovangra [49]1 year ago
8 0

Explanation:

According to orbital hybridization theory, in ethene the C atoms are using sp2 hybrid orbitals. This leaves one unhybridized p orbital available to form the second bond (Pi bond) in the C-C double bonds.

One sp2 hybrid orbital from each C atom overlap to form one sigma bond between the two carbon atoms. One p orbital from each C atom then overlap to form a pi bond that completes the C-C double bond. The H atoms are bonded via sigma bonds to the C atoms when the H s orbitals overlap with the remaining sp2 hybrid orbitals from each C.

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29.4 mL of an CH3COOH solution were titrated with 18.5 mL of a 0.0175 m LiOH solution to reach the equivalence point. What is th
vodomira [7]

Answer:

0.11 M

Explanation:

The computation of molarity of the CH3COOH solution is shown below:-

M_2 = \frac{M_1V_1}{V_2}

Where,

M_1 = Molarity of LioH = 0.0175

V_1 = Volume of LioH = 18.5 ml

V_2 = Volume of CH3COOH = 29.4 ml

Now, we will put the values into the formula

= \frac{0.0175\times 18.5}{29.4}\\\\ = \frac{0.32375}{29.4}

Which gives result

= 0.011 M

Therefore for computing the molarity of the CH3COOH solution we simply applied the above formula.

5 0
3 years ago
Scientists have established that some of the layers of Earth are solid, some are liquid, and some are semi-fluid. Explain how th
tigry1 [53]
They can determine the liquid layers and solid layers by seismic waves
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7 0
4 years ago
Read 2 more answers
how much alum product would be lost to the crystallization solution if you had 42.5 ml of solution after filtration and the solu
JulsSmile [24]

Answer:

Total amount of alum lost = 0.5122 grams

Explanation:

Let the total volume of the solution be 100 mL

In 100 mL of solution, there is 2.63 gram of alum.

Out of this 100 mL solution, 42.5 mL is remaining.

Amount of alum in 42.5 mL solution is

\frac{42.5}{100} * 2.63 = 1.117\\ grams

Now the amount of alum lost is 2.63 -1.117 = 0.5122 grams

3 0
3 years ago
A magnet was placed near a pile that contained both iron and sulfur. The magnet was moved gradually closer to the pile. As it ne
irina1246 [14]

The question is incomplete, the complete question is;

A magnet was placed near a pile that contained both iron and sulfur. The magnet was moved gradually closer to the pile. As it neared the pile, the magnet started attracting small pieces of iron from the pile. Which of these statements best describes the contents of the pile?

F. It is a homogeneous mixture of iron and sulfur.   G. It is a heterogeneous mixture of iron and sulfur.   H. It is a compound that contains both iron and sulfur.   I. It is a compound that can be separated by magnetism.

Answer:

G. It is a heterogeneous mixture of iron and sulfur.

Explanation:

A heterogeneous  mixture is one that does not have a uniform composition throughout.

We must recall that a mixture is any combination of substances that do not chemically react together and are separable by physical means.

Having said this, it is clear that I can separate the iron from sulphur by simple magnetic (physical) means. Hence,  it is a heterogeneous mixture of iron and sulfur.

4 0
3 years ago
Read 2 more answers
PLZ HELP ...The half-life of polonium-218 is 3.0 minutes. If you start with 40.0 g, how long will it be before only 4.5 g remain
Elanso [62]

Answer:-  9.4 minutes.

Solution:- Radioactive decay obeys first order reaction kinetics and the equation used to solve this type of problems is:

lnN=-kt+lnN_0

where, k is decay constant and t is the time. N_0 is the initial amount of the radioactive substance and N is the remaining amount.

Since the value of decay constant is not given, so we need to calculate it first from given half life by using the formula:

k=\frac{0.693}{t_1_/_2}

where t_1_/_2 stands for half life.

Given half life is 3.0 minutes.

So, k=\frac{0.693}{3.0min}

k=0.231min^-^1

Let's plug in the values in the first order reaction equation and solve it for t.

ln4.5g=-0.231min^-^1(t)+ln40.0g

It could also be written as:

ln(\frac{4.5g}{40.0g})=-0.231min^-^1}

-2.18=-0.231min^-^1}

t=\frac{-2.18}{-0.231min^-^1}

k = 9.4 min

So, the radioactive substance would take 9.4 minutes to decay from 40.0 grams to 4.5 grams.

4 0
3 years ago
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