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3 years ago

Compound that contains a terminal carbonyl?

Chemistry

1 answer:

malfutka [58]3 years ago

7 0

Many compunds have a terminal carbonyl

Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.

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Which group on the periodic table would have zero electronegativity because they have a full octet?

bija089 [108]

The group on the periodic table that would have 0 electronegativity due to the fact that their valence shell is full, i.e, have a full octet would be the inert or noble gases. They have a total of 8 electrons in their valence shell and are thus inert and cannot strongly attract electrons toward itself, from neighbouring atom electrons as it does not need to.

3 0

3 years ago

A gas at a constant volume has an initial temperature of 300.0 K and an initial pressure of 10.0 atm. What pressure does the gas

harkovskaia [24]

$\frac{500}{300} times 10=16.667 atm$

4 0

2 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

2 years ago

How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?

LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

$0.500 L \times \frac{0.360mol}{L} = 0.180 mol$

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

$0.180 mol \times \frac{110.98g}{mol} = 20.0 g$

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

You can learn more about solutions here: brainly.com/question/2412491

4 0

1 year ago

Briefly explain the difference between hard water and soft water.

melisa1 [442]

Answer:

I HOPE THE ABOVE INFORMATION WILL HELP YOU A LOT.

HAVE A NICE DAY.

3 0

3 years ago