Answer:
-191.7°C
Explanation:
P . V = n . R . T
That's the Ideal Gases Law. It can be useful to solve the question.
We replace data:
2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°
(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°
T° = 81.3 K
We convert T° from K to C°
81.3K - 273 = -191.7°C
Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
Concentration is decreased to 1.56 % which means that 0.0156 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 0.0156
Thus,
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
Where, k is rate constant
So,
<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>
Answer:
4.56 X 10^ -4 g/mL
Explanation:
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.
(7.6 X10^-4 gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g
this is dissolved )in 10 m L=45.6 X 10^-4 g/ 10
4.56 X 10^ -4 g/mL
check
6/10 =0.6
4.56/7.6 = o.,6
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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