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nirvana33 [79]
3 years ago
8

Label each carbon atom with the appropriate geometry. Bin 1 points to a carbon bonded to a double bonded carbon and single bonde

d to two hydrogens. Bin 2 points to a carbon double bonded to a carbon and single bonded to a carbon and one hydrogen. Bin 3 is a carbon single bonded to two carbons and single bonded to two hydrogens. Bin 4 is the same as bin 3. Bin 5 is a carbon triple bonded to a carbon and single bonded to a carbon. Bin 6 is triple bonded to a carbon and single bonded to a hydrogen.
Chemistry
1 answer:
N76 [4]3 years ago
7 0

Answer:

Bin 1 points to a carbon bonded to a double bonded carbon and single bonded to two hydrogens. --- trigonal planar, tetrahedral

Bin 2 points to a carbon double bonded to a carbon and single bonded to a carbon and one hydrogen.------- trigonal planar, tetrahedral

Bin 3 is a carbon single bonded to two carbons and single bonded to two hydrogens. ----- tetrahedral, tetrahedral

Bin 4 is the same as bin 3.--------tetrahedral, tetrahedral

Bin 5 is a carbon triple bonded to a carbon and single bonded to a carbon.---- linear, tetrahedral

Bin 6 is triple bonded to a carbon and single bonded to a hydrogen.---linear, tetrahedral

Explanation:

A single C-C or C-H bond is in a tetrahedral geometry, the carbon atom is bonded to four species with a bond angle of 109°.

A C=C bond is trigonal planar with a bond angle of 120°.

Lastly, a C≡C bond has a linear geometry with a bond angle of 180° between the atoms of the bond.

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The structure of Methanol (CH₃OH) is shown below. This structure contains 2 lone pair of electrons on oxygen (highlighted red). Electronic configuration of oxygen is,

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3 years ago
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3 0
3 years ago
The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.
Vladimir [108]

Answer:

The molecular formula = C_{6}H_{6}

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,  

n=\frac{m}{M}

Using ideal gas equation as:

PV=\frac{m}{M}RT

where,  

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15

M=78.31\ g/mol

The empirical formula is = CH

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,  

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = C_{6}H_{6}

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3 years ago
In 1911, Ernest Rutherford tested the atomic model existing at the time by shooting a beam of alpha particles (42He, helium nucl
STatiana [176]

Answer:

At the time of Rutherford's experiment, the accepted model for the atom was the Thomson plum-pudding model of the atom, in which the atom consists of a "sphere" of positive charge distributed all over the sphere, with tiny negative particles (the electrons) inside this sphere.

In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

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- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

5 0
3 years ago
What particle decay is this? 210 83 Bi→210 84 Po ​
sesenic [268]

Answer:

Beta emission

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When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.

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