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2 years ago

What type of bond is made between amino acids in the ribosome?

Chemistry

1 answer:

sleet_krkn [62]2 years ago

8 0

Answer:

Peptide bonds form between the amino group of the amino acid attached to the A-site tRNA and the carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide bond is catalyzed by peptidyl transferase, an RNA-based enzyme that is integrated into the 50S ribosomal subunit.

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Please answer this question thank you ^^

alexgriva [62]

Answer:

its number 1 sp^3

Explanation:

3 0

1 year ago

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What units should they have used in order to make the correct conversion

grandymaker [24]

We need to see your problem/question you’re talking about to answer it

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3 years ago

A dilute solution of bromine in carbon tetrachloride behaves as an ideal-dilute solution. The vapour pressure of pure CCl4 is 33

ANEK [815]

Explanation:

The given data is as follows.

Vapour pressure of pure $CCl_{4}$ = 33.85 Torr

Temperature = 298 K

Mole fraction of $Br_{2}$ = 122.36 torr

Therefore, calculate the vapor pressure of $Br_{2}$ as follows.

Vapour pressure of $Br_{2}$ = mole fraction of $Br_{2}$ x K of $Br_{2}$

= 0.050 x 122.36 Torr

= 6.118 Torr

So, vapor pressure of $Br_{2}$ is 6.118 Torr .

Now, calculate the vapor pressure of carbon tetrachloride as follows.

Vapour pressure of $CCl_{4}$ = mole fraction of $CCl_{4}$ x Pressure of $CCl_{4}$

= (1 - 0.050) × 33.85 Torr

= 32.1575 Torr

So, vapor pressure of $CCl_{4}$ is 32.1575 Torr .

Hence, the total pressure will be as follows.

= 6.118 Torr + 32.1575 Torr

= 38.2755 Torr

Therefore, composition of $CCl_{4}$ = $\frac{32.1575 Torr}{38.2755 Torr}$

= 0.8405

Composition of $CCl_{4}$ is 0.8405 .

And, composition of $Br_{2}$ = $\frac{6.118 Torr}{38.2755 Torr}$

= 0.1598

Composition of $Br_{2}$ is 0.1598 .

6 0

3 years ago

Use the IUPAC nomenclature rules to give the name for this compound - Al2S3.

jonny [76]

Aluminium Sulfide
According to rules the positive specie is named first and the negative specie is named last.

6 0

2 years ago

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What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer

Zepler [3.9K]

$d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}$

8 0

3 years ago