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kvv77 [185]
2 years ago
10

Which functional group is found in an amine?

Chemistry
2 answers:
kirill [66]2 years ago
8 0

Answer:

N is single bonded to R, and single bonded twice to H.

Explanation:

The functional group present in an amine will not be complete without mentioning the nitrogen group present in the compound.

Amines are organic compounds that show appreciable basicity. They are regarded as the derivatives of ammonia where one or two or the three hydrogen atoms has/have been replaced been replaced by alkyl or aryl groups.

  • Amines are classified as primary, secondary or tertiary based on the number of substituted hydrogen atoms they carry.
  • The nitrogen bonded to the R - aryl group and bonded to hydrogen is a good description of the functional group.
natima [27]2 years ago
8 0

Answer:

b

Explanation:

because i said so brudda

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C is the answer I took a test with the same answer
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SOICHIOMETRY HOW MANY MOL OF NITROGEN DIOXIDE ARE REQUIRED TO PRODUCE 2.75 MOL NITRIC ACID? 3NO2+ H20→ 2HNO3+ NO
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2 years ago
A compound of mercury and oxygen is heated in order to decompose the compound. A 4.08 grams sample of mercury oxide upon heating
arlik [135]

Answer:

HgO (empirical formula)

Explanation:

4.08 - 3.78 = 0.3g (oxygen)

(\frac{4.08}{201})   \:  \:  \:  (\frac{0.3}{16} )

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7 0
2 years ago
What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point
Darya [45]

<u>Answer:</u> The amount of energy absorbed by water is 5390 Calories

<u>Explanation:</u>

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

q=m\times L_{vap}

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

L_{vap} = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

q=10g\times 539Cal/g=5390Cal

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6 0
3 years ago
A sample of gas initially has a volume of 2.25 L at 350 K and a pressure of 1.75 atm. What will be sample pressure if the volume
IRINA_888 [86]

Answer:

8.44 atm

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 2.25 L

Initial temperature (T₁) = 350 K

Initial pressure (P₁) = 1.75 atm

Final volume (V₂) = 1 L

Final temperature (T₂) = 750 K

Final pressure (P₂) =?

The final pressure of the gas can be obtained as illustrated below:

P₁V₁/T₁ = P₂V₂/T₂

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3.9375 / 350 = P₂ / 750

Cross multiply

350 × P₂ = 3.9375 × 750

350 × P₂ = 2953.125

Divide both side by 350

P₂ = 2953.125 / 350

P₂ = 8.44 atm

Thus, the final pressure of the gas is 8.44 atm.

7 0
3 years ago
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