Cl-35, as the atomic mass of Chlorine (35.45) is closer to the number 35 than to the number 37. A higher abundance of CL-35 isotope would have caused the atomic number (which is an average of the values of all isotopes of a substances taking relative abundance into consideration) to decrease from 36, which would appear to be the average.
Answer: c = 710 J/kg°C or 0.71 J/g°C
Explanation: Heat is expressed in the formula Q = mc∆T. Derive to find the specific heat c. So the formula will become c = Q / m∆T
c = Q / m∆T
= 42600 J / 2 kg ( 55°C - 25°C )
= 710 J /kg°C
Or can be expressed by converting kg to g.
c = 0.71 J /g°C
Answer:
1. Lysine
2. Aspartic acid
3. Serine
4. Alanine
5. Tryptophan
Explanation:
Amino acids are biomolecules that contain two functional groups and one R side chain. The two functional groups are: carboxyl group and amino group.
The α-amino acids are the amino acids in which the two functional groups and the R side chain are attached to the α-carbon of the amino acid. They are total 22 α-amino acids.
1. A basic amino acid: Lysine is a positively charged, polar basic amino acid with a lysyl side chain.
2. An acidic amino acid: Aspartic acid is a negatively charged, polar acidic amino acid with an acidic carboxymethyl group.
3. A neutral polar amino acid: Serine is a polar and neutral amino acid with a hydroxymethyl group.
4. A non-polar aliphatic amino acid: Alanine is an aliphatic, nonpolar and neutral amino acid with a methyl side chain.
5. An aromatic amino acid: Tryptophan is an aromatic, nonpolar and neutral amino acid with an indole side chain.
To solve this problem, the dilution equation (M1 x V1 = M2 X V2) must be used. The given values in the problem are M1= 12.0 M, V1= 30 mL, and M2= 0.160 M. To solve for V2,
V2=M1xV1/M2
V2= (12x30)/0.16
V2= 2,250 mL.
The correct answer is 2.25 L.
