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Rom4ik [11]
3 years ago
8

How many codons are needed to specify 3 amino acids?

Chemistry
1 answer:
Nataliya [291]3 years ago
5 0
There need be 9 amino acids, as each amino acid is composed of 3 codons.
Hope this helps!

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How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
trapecia [35]

Answer:

9.9 ml of 0.200M NH₄OH(aq)

Explanation:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

?ml of 0.200M NH₄OH(aq) reacts completely with  12ml of 0.550M FeCl₃(aq)

1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution

1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters  

5 0
2 years ago
Ammonium phosphate nh43po4 is an important ingredient in many fertilizers. it can be made by reacting phosphoric acid h3po4 with
Tom [10]
The reaction between phosphoric acid and ammonia that produces ammonium phosphate can be written as follows:
3NH3 + H3PO4 ..................> (NH4)3PO4

From the periodic table:
molar mass of nitrogen = 14 grams
molar mass of hydrogen = 1 grams
molar mass of oxygen = 16 grams
molar mass of phosphorus = 30.9 grams

based on this:
molar mass of 3NH3 = 3 (14 + 3(1)) = 51 grams
molar mass of H3PO4 = 3(1) + 30.9 + 4(16) = 97.9 grams
molar mass of  (NH4)3PO4 = 3 (14 + 4(1)) + 30.9 + 4(16) = 54 + 30.9 + 64
                                            = 148.9 grams

Therefore, 97.9 grams of phosphoric acid is required to produced 148.9 grams of ammonium phosphate.
Thus, to know the mass of ammonium phosphate produced from 4.9 grams of phosphoric acid, we will simply use cross multiplication as follows:
amount of produced ammonium phosphate = (4.9 x 148.9) / 97.9 = 7.45 g
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3 years ago
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the surface tension of H20 is 72 dynes/cm at 25°C

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I think it would be African plate
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algol [13]
13. D
14. A
15. C
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