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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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dezoksy [38]
Hardness = 75ppm = 75 mg / L

volume = 50 mL =  0.05 L

So, applying hardness formula:

Hardness = mass / volume

so, mass =  hardness x volume = 75 x 0.05 =  3.75 mg = 0.00375 g

So, moles of CaCO3 = moles of Ca2+ ions = mass / molar mass of CaCO3 =  0.00375 / 100.06  =  0.00003747751 moles 

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The person was travelling 100 kph (kilometers per hour). Hope this helps :)
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3 years ago
A cubic meter of air was sampled at 1.0 ATM and 40 C.
ladessa [460]

The amount of nitrogen oxide in the air sample is determined as 7.01 ppb.

<h3>Amount of Nitrogen oxide in the air</h3>

The amount of Nitrogen oxide (NO2) in the air in parts per billion (ppb) is calculated as follows;

12 micrograms of NO2 = 12 μg = 12 x 10⁻⁶ g

PV = nRT

V = nRT/P

Where;

  • V is volume of NO2 gas
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n =  12 x 10⁻⁶/46

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<h3>Amount of NO2 In parts per billion</h3>

= (Volume of NO2)/(volume of air) x 10⁹

= (7.01 x 10⁻⁹)/(1) x  10⁹

= 7.01 ppb

Thus, the amount of nitrogen oxide in the air sample is determined as 7.01 ppb.

Learn more about nitrogen oxide here: brainly.com/question/13629381

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2 years ago
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The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
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Compare and contrast the compositions of binary ionic and binary molecular compounds
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