Final volume is 400 mL
<span>The moles in MgSO4 is 0.00788 </span><span>mL
</span>
The new concentration is 0.197
The answer I believe is 3.340kj.
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
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Answer:
If two tectonic plates collide, they form a convergent plate boundary. Usually, one of the converging plates will move beneath the other, a process known as subduction. ... As the sinking plate moves deeper into the mantle, fluids are released from the rock causing the overlying mantle to partially melt.
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
