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miss Akunina [59]

2 years ago

10

Why there are only 20 amino acids, although a amino acid is coded by a single codon with three bases. That means by permutation,

3*3*3=27 amino acids must exist ??? ...?

1 answer:

snow_lady [41]2 years ago

5 0

In nature there are many more variations amino acids than the simple 20 found in humans. However, when analyzing the human genome sequence, there is a code for all 64 permutations (4^3), only some of them share amino acids. This is a safe-guard against mutations of one or two nucleotides. For example, the amino acid Alanine is coded with four different nucleotide sequences: GCA, GCC, GCG, GCU. Also some amino acids code the same like UUU &UUC

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neonofarm [45]

Answer:

The answer to your question is 0.10 M

Explanation:

Data

Molarity = ?

mass of Sucrose = 125 g

volume = 3.5 l

Formula

Molarity = moles / volume

Process

1.- Calculate the molar mass of sucrose

C₁₂H₂₂O₁₁ = (12 x 12) + (1 x 22) + (16 x 11)

= 144 + 22 + 176

= 342 g

2.- Convert the mass of sucrose to moles

342 g of sucrose ------------------- 1 mol

125 g of sucrose -------------------- x

x = (125 x 1) / 342

x = 0.365 moles

3.- Calculate the molarity

Molarity = 0.365 / 3.5

4.- Result

Molarity = 0.10

5 0

2 years ago

Compare an ionic and a molecular compound in terms of how each is formed

GREYUIT [131]

Answer:

Molecular compounds are pure substances formed when atoms are linked together by sharing of electrons while ionic compounds are formed due to the transfer of electrons.

Molecular compounds are made due to covalent bonding while ionic compounds are made due to ionic bonding.

Explanation:

8 0

2 years ago

A man is 72 inches tall, what is his height in meters? round to three significant figures.

Artist 52 [7]

Answer:

1.829 meters is the answer

Explanation:

5 0

3 years ago

Read 2 more answers

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0

2 years ago