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2 years ago

Draw a lewis structure for ketene, c2h2o, which has a carbon–carbon double bond

Chemistry

1 answer:

MrMuchimi2 years ago

8 0

Visual representation of covalent bonding indicating the valence shell electrons in the molecule, lines represents the shared pair of electron and pair of electrons that are not involved in bonding are represented as dots(lone pairs) are known as Lewis structures.

Compound formation takes place in order to complete the octet of each element that is according to octet rule, each atom forms bond with other atom in order to complete their octet that is to get eight electrons in its valence shell and attain stability.

An organic compound of the form $R^{'}R^{''}C=C=O$ is known as ketene.

The given ketene is $C_2H_2O$ .

The number of valence electron of:

$C = 4$

$H = 1$

$O = 6$

The number of valence electrons in $C_2H_2O$ = $2\times 4+2\times 1+1\times 6 =16$

2 electrons are involved in each single bond between carbon and hydrogen and 4 electrons are involved in each double bond formed between carbon-carbon and carbon-oxygen. Hence, the total number of electrons involved in bond formation are 12 and rest 2 pair of electrons are present on oxygen as lone pair of electrons.

Therefore, the attached image is the Lewis structure of $C_2H_2O$ .

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Plz help. I just need someone to check answer. :(

My name is Ann [436]

These substances can be separated by distillation, so your answer is A.

5 0

2 years ago

At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepare

OleMash [197]

Answer:

$P_{CCl_4}=52.43torr$

Explanation:

Hello there!

In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

$P_{CCl_4}=x_{CCl_4}P_{CCl_4}^{vap}$

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

$x_{CCl_4}=\frac{140/153.81}{140/153.81+67.1/100.21}=0.576$

Therefore, the partial pressure of chloroform turns out to be:

$P_{CCl_4}=0.576*91torr\\\\P_{CCl_4}=52.43torr$

Regards!

3 0

2 years ago

Calculate number of moles of 4.4g of CO2 and 5.6L of NH3 . Pls help . Its urgent

kupik [55]

Moles of CO₂ = mass / molecular weight
Moles of CO₂ = 4.4 / (12 + 16 x 2)
Moles of CO₂ = 0.1 mol

Each mole of gas occupies 22.4 L at STP. Therefore,
Moles of NH₃ = 5.6 / 22.4
Moles of NH₃ = 0.25 mol

4 0

3 years ago

What is the boiling point of water at an elevation of 1.55×104 ft ?

horsena [70]

The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)

4 0

2 years ago

Consider the following system at equilibrium:

alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0

2 years ago