Answers:
1) - COOH
2) carboxyl group
The organic acids contain the carboxyl group: - COOH.
Fats are structures composed by fatty acids. Fatty acids are long aliphatic chains with a carboxyl group at the end.
Aminoacids are chains that contains amine group (NH2) at one end and the carboxyl group at the other end.
Then both aminoacids and fatty acids contain the carboxyl group (-COOH) at one end.
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Among the choices given, the correct answer is the first option. Substance X most likely is a crystal, and substance Y most likely is a liquid. Substance X having a fixed volume describes a crystal because crystals occupy a certain volume. Substance Y is a liquid because liquids can still be compressed further in order to attain a more packed <span>structure.</span>
Answer:
6.564×10¹⁶ fg.
Explanation:
The following data were obtained from the question:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt in fg =?
Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt =?
Mass of salt = (Mass of beaker + salt) – (Mass of beaker)
Mass of salt = 142.54 – 76.9
Mass of salt = 65.64 g
Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:
Recall:
1 g = 1×10¹⁵ fg
Therefore,
65.64 g = 65.64 g × 1×10¹⁵ fg / 1g
65.64 g = 6.564×10¹⁶ fg
Therefore, the mass of the salt is 6.564×10¹⁶ fg.
Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

The molar mass will be thus:

Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

The percent composition of Carbon is thus 12.7 %.