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1 year ago

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18. give the product for the following reactions sequence. a) 3-pentanone b) 2-pentanone c) propanone d) 2-butanone e) methyl pr

opanoate 19. give the product for the following reactions sequence. a) ch3ch2cocl b) ch3ch2cho c) ch3co2ch3 d) ch3ch2cooch3 e) ch3ch2ch2cl 20. give the product for the following reactions sequence. a) ch3ch2cocl b) ch3ch2conhch3 c) ch3conhch2ch3 d) ch3ch2cooch3 e) ch3ch2ch2cl

1 answer:

serg [7]1 year ago

5 0

The product for the following reaction are :

2 - butanone
CH₃CH₂COCl
CH₃CH₂CONHCH₃

T carboxylic acid is an organic compound. the functional group of carboxylic acid is carboxy. general formula is given as : R - COO⁻.

The reactions are given as follows :

1) the reduction of carboxylic acid into ketone with the organometallic compound is given as follows :

CH₃CH₂COOH + CH₃Li ------> CH₃CH₂COCH₃

2 butanone

2) the reaction of carboxylic acid with SOCl₂ is given as follows :

CH₃CH₂COOH + SOCl₂ -----> CH₃CH₂COCl

3) the reaction of carboxylic acid with CH₃NH₂ is given as follows

CH₃CH₂COOH + CH₃NH₂ -----> CH₃CH₂CONHCH₃

Thus, The product for the following reaction are :

2 - butanone
CH₃CH₂COCl
CH₃CH₂CONHCH₃

To learn more about carboxylic acid here

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A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

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3 years ago

What is the answer to number one

Shalnov [3]

2 the answer is 2 because 1.0 * 2 =2

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3 years ago

Calculate the wavelength of a red light with a frequency of 4.61 x 1012 Hz.

Naddika [18.5K]

Answer:

4665.32

Explanation:

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2 years ago

At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume

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Answer: Volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

Explanation:

Given: $T_{1} = 35^{o}C = (35 + 273) K = 308 K$ , $V_{1}$ = 256 mL,

$P_{1}$ = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

$T_{1} = 22^{o}C = (22 + 273) K = 295 K$ , $P_{2} = 1.25 atm$

Formula used to calculate volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL$

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

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2 years ago

love history [14]

Answer : The molar concentration of sucrose in the tea is, 0.0549 M

Explanation : Given,

Mass of sucrose = 3.765 g

Volume of solution = 0.200 L

Molar mass of sucrose = 342.3 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}\times \text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{3.765g}{342.3g/mole\times 0.200L}=0.0549mole/L=0.0549M$

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M

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3 years ago