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3 years ago

What are the derivatives of alkanoic acid?

Chemistry

1 answer:

Nuetrik [128]3 years ago

4 0

The answer to your question is,

Carboxylic acids themselves, carboxylates (deprotonated carboxylic acids), amides, esters, thioesters, and acyl phosphates.

-Mabel <3

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How many grams of water will be produced when 1.6 moles of ethanol (CH3CH2OH) are burned completely? Enter a number only (no uni

AfilCa [17]

Answer:

Explanation:

The reaction that takes place is:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

First we <u>convert moles of ethanol to moles of water</u>:

1.6 mol ethanol * $\frac{3molH_2O}{1molEthanol}$ = 4.8 mol H₂O

Then we <u>convert moles of water to grams of water</u>, using its molar mass:

4.8 mol H₂O * 18 g/mol = 86.4 g

So 84.6 grams of water will be produced.

4 0

2 years ago

What is the differnce between the atomic number and the mass number of an element?

MrRissso [65]

Atomic number refers to the number of protons within the nuclei of an element's atoms (and therefore the number of electrons) while mass number refers to the number of protons and neutrons within the nucleus.

8 0

3 years ago

Each of the diagrams below shows an electrically neutral atom. fill in the missing number of protons (n) neutrons (and), electro

vova2212 [387]

What is the diagram?

7 0

3 years ago

11. A new compound has just been developed. Its formula is:

jarptica [38.1K]

Answer:

Explanation:

G, H, O, N, Na, P

There are 6 different elements listed, with O (oxygen) showing up twice.

7 0

3 years ago

what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c

slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

$Q = m \times C \times \bigtriangleup T$

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

$C = \frac{Q}{m \times \bigtriangleup T}$

Let us plug in the values in above equation.

$C = \frac{300J}{267g \times 12 C}$

$C = \frac{300J}{3204 g C}$

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0

3 years ago