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Ede4ka [16]
1 year ago
7

Question 1: Which of the following statement is false about conjugated systems?

Chemistry
1 answer:
valkas [14]1 year ago
8 0

Answer:

A conjugated system can only contain two alkenes: False

B. A cumulated diene is less stable than an isolated diene.

False

Explanation:

A. A conjugated system is more stable than an unconjugated system.: True

Due to resonance a conjugated system is more stable.

B. The s-trans isomer is favored over the s-cis isomer.: True

Due to repulsion in s-cis isomer they are less stable.

C. A conjugated system has multiple resonance structures. True

This is the reason they are stable.

D. A conjugated system can only contain two alkenes: False

A conjugated system may have may alternate double and triple bonds.

2) A trans alkene releases less heat upon hydrogenation than a cis alkene.

True

Trans alkenes are more stable than cis alkene. Thus heat of hydrogenation of trans alkene is less than cis alkene.

B. A cumulated diene is less stable than an isolated diene.

False

C. Multiple products can always form when treated a conjugated diene with HBr.  True

There can be 1,2 or 1,4 addition

D. Conjugated systems arise only between sp2-hybridized atoms.

True

Conjugation allows resonance which is possible in sp2 systems

E. Conjugated double bonds involve unhybridized p-orbitals aligned with each other

True

The double bond is formed as pi bond due to side ways overlapping of unhybridized p orbitals

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Suppose that the microwave radiation has a wavelength of 12.4 cm . How many photons are required to heat 255 mL of coffee from 2
vivado [14]

Answer:

Explanation:

wavelength λ = 12.4 x 10⁻² m .

energy of one photon = h c / λ

= 6.6 x 10⁻³⁴ x 3 x 10⁸ /  12.4 x 10⁻²

= 1.6 x 10⁻²⁴ J .

Let density of coffee be equal to density of water .

mass of coffee = 255 x 1 = 255 g

heat required to heat up coffee = mass x specific heat x rise in temp

= 255 x 4.18 x ( 62-25 )

= 39438.3 J  .

No of photons required = heat energy required / energy of one photon

= 39438.3 / 1.6 x 10⁻²⁴

= 24649 x 10²⁴

= 24.65 x 10²⁷ .

5 0
1 year ago
A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur
Marat540 [252]

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

P1 V1 / T1 = P2 V2/ T2

Derive to find V2

V2 = P1 V1 T2 / T1 P2

Convert temperatures in K

T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

Substitute the values.

4 0
2 years ago
A chemist wants to extract copper metal from copper chloride solution. The chemist places 1.50 grams of aluminum foil in a solut
Lisa [10]

Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Explanation: 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.

Thus 54 g of of aluminium react with 270 g of copper chloride.

1.50 g of aluminium react with=\frac{270}{54}\times 1.50=7.5 gof copper chloride.

3 moles of copper chloride gives 3 moles of copper.

7.5 g of copper chloride gives 7.5 g of copper.

8 0
2 years ago
Read 2 more answers
How many structures are possible for a trigonal bipyramidal molecule with a formula of ax3y2?
ryzh [129]
Check the attached file for the answer.

4 0
2 years ago
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
1 year ago
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