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2 years ago

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In which of the compounds below is there more than one kind of hybridization (sp, sp2, sp3) for carbon? I. CH 3CH 2CH 2CH 3 II.

CH 3CH = CHCH 3 III. CH 2 = CH – CH = CH 2 IV. H – C ≡ C – H

1 answer:

77julia77 [94]2 years ago

8 0

The compounds containing more than one kind of hybridization (sp, sp2, sp3) for carbon is CH3CH = CHCH3.

<u>Explanation</u>:

Orbital hybridization is the idea of blending nuclear orbitals into new hybrid orbitals reasonable for the matching of electrons to form chemical bonds in the valence bond hypothesis.

Here CH3CH = CHCH3 contains a carbon atom joined to more than 3 hydrogen atoms.

In sp – Hybridisation 1 s-orbital and 1 p-orbital are blended to form two sp – hybrid orbitals, having a straight structure. In sp2 – Hybridisation 1 s-orbital and 2 p-orbitals are blended forming three sp2–hybrid orbitals, having a planar triangular structure. In sp3 – Hybridisation 1 s-orbital and 3 p-orbitals are blended forming four sp3–hybrid orbitals having a tetrahedral structure.

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You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

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Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

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1 year ago

Select the number of electrons each atom must gain or lose to have a full valence level. Use the periodic table if you need help

Lady bird [3.3K]

Calcium will loose one electron. Fluorine will gain one electron. Lithium will loose one electron. Argon will not loose any because it already has a full valence level. Aluminium will loose 3 electrons.

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2 years ago

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When working with food, personal items such as a cell phone should be

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Answer:

should be put away in a bag or a pocket away from the food

Explanation:

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2 years ago

A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in

Fittoniya [83]

The molarity (M) of the acid in the lake water is $0.00001M$ .

<u>Explanation:</u>

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

$M_1V_1=M_2V_2$

Here;

$M_1,M_2$ are molarity of acid in the lake water and base solution respectively. $V_1,V_2$ are volume of sample in the lake water and base solution respectively.

Given values are as follows:

$M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml$

Putting these values in above equation :

$M_1V_1=M_2V_2$

$M_1(1000)=(0.100)(0.100)$

$M_1=\frac{(0.100)(0.100)}{1000}$

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1 year ago

An unknown compound with a molar mass of 155.06 g/mol consists of 46.47% c, 7.80% h, and 45.72% cl. find the molecular formula f

Contact [7]

Answer:- Molecular formula of the compound is $C_6H_1_2Cl_2$ .

Solution:- From given information:

C = 46.47%

H = 7.80%

Cl = 45.72%

First of all we find out the empirical formula from given percentages. We divide the given percentages by their respective atomic masses to calculate the moles:

$C=\frac{46.47}{12.01}$ = 3.87

$H=\frac{7.80}{1.01}$ = 7.72

$Cl=\frac{45.72}{35.45}$ = 1.29

Now, we divide the moles of each by the least one of them. Least one is Cl as it's moles are least as compared to the moles of C and H. So, let's divide the moles of each by 1.29.

$C=\frac{3.87}{1.29}$ = 3

$H=\frac{7.72}{1.29}$ = 6

$Cl=\frac{1.29}{1.29}$ = 1

So, the empirical formula of the compound is $C_3H_6Cl$ .

Empirical formula mass = 3(12.01) + 6(1.01) + 1(35.45)

= 36.03 + 6.06 + 35.45

= 77.54

To calculate the number of formula units we divide molar mass by empirical formula mass.

number of empirical formula units = $\frac{155.06}{77.54}$

= 2

So, the molecular formula would be two times of empirical formula that is, $C_6H_1_2Cl_2$ .

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2 years ago

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