Okay so,
1) Translation- show the RNA strand attatching to a DNA strand with the complimentary base pairs. introns are spliced
2) mRNA leaves the cell and joins with a ribosome
3) Transcription - tRNA (clover shaped) reads each codon (triplets) which each code for an amino acid. The stop codons on the end tell the tRNA that the chain is finished
4) the sequence forms the primary structure (all peptide bonds) which determines the shape of the secondary (hyrdogen and peptide) and hence determines the shape of the tertiary structure of a protein (ionic, hydrogen, disulfide bridges and hydrophibic interactions)
Hope this helps :)
Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol
Moles=14.5g / 58.14g/mol
=0.249
Therefore there are approx 0.249 moles in a 14.5g sample of C4H10
Answer:
K₂Cr₂O₇(s) ⇒ 2 K⁺(aq) + Cr₂O₇²⁻(aq)
Explanation:
Potassium dichromate (K₂Cr₂O₇) is a strong electrolyte, that is, when dissolved in water (the medium), it dissociates in cation potassium (K⁺) and anion dichromate (Cr₂O₇²⁻). The balanced dissociation equation is:
K₂Cr₂O₇(s) ⇒ 2 K⁺(aq) + Cr₂O₇²⁻(aq)
Answer:
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
Explanation:
The equation for the reaction of trimethylamine when it is dissolved in water is:
C₃H₉N + H₂O ⇄ C₃H₉NH⁺ + OH⁻ Kb
1 mol of trimethylamine catches a proton from the water in order to produce trimethylamonium.
It is a base, because it give OH⁻ to the medium
Expression for Kb (Molar concentration)
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
