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-Dominant- [34]
1 year ago
13

Part B

Chemistry
1 answer:
sashaice [31]1 year ago
7 0

The anticodons corresponding to the codons on the mRNA (from part A) is 5' CGA - AAA - GUU 3'.

<h3>What are anticodons?</h3>

Anticodons are nucleotide sequences on tRNA molecules that are complementary to the codons found on mRNA molecules.

The anticodons on tRNA molecules determine the amino acid that is carried by the tRNA.

Just like codons, anticodons occur in triplets of nucleotide sequences.

Considering the codons on the mRNA molecule:

3’ GCT | TTT | CAA | AAA ’5

The complementary anticodon will be:

5' CGA - AAA - GUU 3'

Learn more about anticodons at:brainly.com/question/28067314

#SPJ1

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Determine the density of gold if a 475.09g sample occupy a space of 24.6cm^3
Ira Lisetskai [31]

Answer:

The answer is

<h2>19.31 g/cm³</h2>

Explanation:

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density =  \frac{mass}{volume}

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The density of the gold is

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Hope this helps you

6 0
3 years ago
If you start with 13 mole H2O and 50 grams of O2 what is the percent yield of H202 if 100 grams of H202 is actually made?
dmitriy555 [2]

Answer:

94.1 %

Explanation:

We firstly determine the equation:

2H₂O + O₂ →  2H₂O₂

2 moles of water react to 1 mol of oxygen in order to produce 2 moles of oxygen peroxide.

We convert the mass of oxygen to moles:50 g . 1mol /32g = 1.56 mol

Certainly oxygen is the limiting reactant.

2 moles of water react to 1 mol of oxygen.

13 moles of water may react to 13/2 = 6.5 moles. (And we only have 1.56)

As we determine the limiting reactant we continue to the products:

1 mol of O₂ can produce 2 moles of H₂O₂

Then 1.56 moles of O₂ will produce (1.56 . 2) = 3.125 moles

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Percent yield = (Yield produced / Theoretical yield) . 100

(100g / 106.25 g) . 100 = 94.1 %

3 0
3 years ago
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