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Dima020 [189]
3 years ago
11

Specify the functional groups by letters (A to G) as given in the list below. Use only one letter per box. If there are fewer th

an three functional groups, leave the appropriate number of answer boxes empty. If a functional group is present MORE than once, provide ONLY ONE entry for these groups.)
CH3-HC(NH2)-CO-OH

(a) hydroxyl group in alcohol
(b) amino group
(c) aldehyde group
(d) ketone group
(e) carboxyl group
(f) ester group
(g) amide group
Chemistry
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

(b) amino group

e) carboxyl group

Explanation:

The functional group in an organic compound distinguishes it from other compounds. It is usually the site where chemical reactions takes place.

       CH₃-HC(NH₂)-COOH

There are two functional groups in this organic compoud:

  • The NH₂ group is called the amino group
  • The COOH is a carboxyl group

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There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen
alekssr [168]

Answer:

2.765amu is the contribution of the X-19 isotope to the weighted average

Explanation:

The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:

X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21

The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:

19.00amu * 0.1455 =

2.765amu is the contribution of the X-19 isotope to the weighted average

8 0
3 years ago
How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?
PIT_PIT [208]
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
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7 0
4 years ago
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Wewaii [24]
Answer: A) Water Explanation: Water molecules are polar molecules
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A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
ICI(g) + Cl₂(g) → ICl3(s). The AG(f) of ICI(g) is -5.5 kJ/mol and the
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Answer:

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Explanation:

3 0
2 years ago
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