Choose the option below that is a characteristic of ketones. a. They contain a carbonyl group that exhibits sp hybridization around the carbon atom.
b. They contain a carbonyl group with a nonpolar carbon-oxygen bond.
c. The functional group of this type of compound must always be on the end of a carbon chain.
d. The functional group of this type of compound must always be in the middle of a carbon chain.

Answers

Answer 1

Answer:

Answer:

Option d.

Explanation:

Ketones contain a carbonyl groups as a functional group, which is a carbon bonded to oxygen with a double bond. In a ketone, the carbon is always bonded to two carbon atoms:

R-C(=O)-R'

The carbon in the carbonyl group has a hybridization sp2 (3 hybrid orbitals with an unhybridized p orbital), where two of the orbitals form sigma (σ) bonds with the other two carbons (R-C-R') and the other hybrid orbital form a sigma bond with the oxygen (C-O). The unhybridized p orbital on the carbon atom is used to form a pi (π) bond with the oxygen, thus forming the double bond (C=O).

The bond of a carbonyl group is polar, because of the difference of the electronegativity between the carbon atom and the oxygen atom.

Hence, from all of the above we can discard the option a, (the carbonyl groups exhibits sp2 hybridization), the option b (carbon-oxygen bond is a bond polar) and the option c (the group must always be in the middle of a carbon chain, the groups that are always in the end, are a aldehyde groups).

Therefore, the correct option is d, the functional group of this type of compound must always be in the middle of a carbon chain.

I hope it helps you!

Answer 2

Answer:

Answer:

d. The functional group of this type of compound must always be in the middle of a carbon chain.

Explanation:

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Round each answer to the number of significant figures in the parentheses.

Answers

Answer:

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Draw all the structural isomers for the molecular formula C3H8O. Be careful not to draw any structures by crossing one line over another; the system needs to know where you intend connections to be between atoms.

Answers

Answer:

The three isomers having the molecular formula $C_(3) H_(8)O$ are drawn in the figure below.

Explanation:

7. What is the molar mass of each of the following elements?a) helium, He(s)
c) potassium, K(s)
b) manganese, Mn(s)
d) boron, B(s)

Answers

Answer:

The molar mass of:

Helium = 4.00 g/mol

Potassium = 39.0983 g/mol

Manganese = 54.94 g/mol.

Boron = 10.81 g / mol

Explanation:

Helium = 4.00 g/mol

Potassium = 39.0983 g/mol

Manganese = 54.94 g/mol.

Boron = 10.81 g / mol

What change would you expect on the rate of the SN2 reaction of 1-iodo-2-methylbutane with cyanide ion if the nucleophile concentration is halved and the alkyl halide concentration is unchanged ?

Answers

Answer:

Rate of reaction will be half of it's initial value

Explanation:

For the given $S_(N)2$ reaction, the rate law is -

$Rate=k[1-iodo-2-methylbutane][CN^(-)]$

Where k is rate constant, [1-iodo-2-methylbutane] is concentration of 1-iodo-2-methylbutane and $[CN^(-)]$ is concentration of $CN^(-)$

Here nucleophile is the $CN^(-)$ ion

Initiallly, $(Rate)_(initial)=k* [1-iodo-2-methylbutane]_(initial)* [CN^(-)]_(initial)$

When concentration of $CN^(-)$ is halved then-

$Rate=k* [1-iodo-2-methylbutane]_(initial)* ([CN^(-)]_(initial))/(2)=((Rate)_(initial))/(2)$

So rate of reaction will be half of it's initial value

Given that e = 9.0 v , r = 98 ω and c = 23 μf , how much charge is on the capacitor at time t = 4.0 ms

Answers

Let charge across the capacitor be Q, current through the circuit be I.
Voltage difference across the resistor = rI
Voltage difference across the capacitor = Q/c
Loop rule: net voltage change through a loop must be zero, so
9 = rI + Q/c. Since I = dQ/dt,
r dQ/dt + Q/c = 9
Solving, Q = 9c (1 - e^(t/rc)). Plug in the numbers from the problem for the numerical answer.

4Ga + 3S2 → 2Ga2S3 1. How many grams of Gallium Sulfide would form if 20.5 moles of Gallium burned?

Answers

Answer:

$m_(Ga_2S_3)=2415.31gGa_2S_3$

Explanation:

Hello,

In this case, for the given chemical reaction, we can notice there is a 4:2 molar ratio between the burned moles of gallium and the yielded moles of gallium sulfide, therefore, we compute them as shown below:

$n_(Ga_2S_3)=20.5molGa*(2molGa_2S_3)/(4molGa)=10.25mol Ga_2S_3$

Then, by using the molar mass of gallium sulfide (235.64 g/mol), we directly compute the grams:

$m_(Ga_2S_3)=10.25mol Ga_2S_3*(235.64gGa_2S_3)/(1molGa_2S_3) \n\nm_(Ga_2S_3)=2415.31gGa_2S_3$

Best regards.

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