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3 years ago

19. Tryptophan is an essential _____ that can be found in the genetic code of humans.

Chemistry

1 answer:

Ivan3 years ago

3 0

The answer is B tryptophan is a amino acid

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Which factors must be equal when a reversible chemical process reaches equilibrium?

Anon25 [30]

The concentration of the reactants and products remain constant. Because the rates of the forward and reverse reaction are equal there is no net change to the amount of reactants or products produced.May 19, 2011

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3 years ago

Who discovered sodium?

Free_Kalibri [48]

Humphry Davy First discovered sodium

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3 years ago

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What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 = [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!

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3 years ago

I need the answer please

dlinn [17]

Answer:

Na+ the answer is A

Explanation:

6 0

3 years ago

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In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign

zloy xaker [14]

Explanation:

Expression for the $v_{rms}$ speed is as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

where, $v_{rms}$ = root mean square speed

k = Boltzmann constant

T = temperature

M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of $v_{rms}$ as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}$

= 498.5 m/s

Hence, $v_{rms}$ for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its $v_{rms}$ speed as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}$

= 524.5 m/s

Therefore, $v_{rms}$ speed for nitrogen is 524.5 m/s.

3 0

3 years ago