Things go sublime when changing automatically into vapor when it is heated, usually forming a solid figure.
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
Answer:
a) Volume of vial= 9.626cm3
b) Mass of vial with water = 62.92 g
Explanation:
a) Mass of empty vial = 55.32 g
Mass of Vial + Hg = 185.56 g
Therefore,

Density of Hg = 13.53 g/cm3

b) Volume of water = volume of vial = 9.626 cm3
Density of water = 0.997 g/cm3

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.
An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given :
The ideal gas equation is given below.
n = PV/RT
n = 86126.25 x 0.0024 / 8.314 x 287
n = 0.622 / molar mass (n = Avogardos number)
Molar mass = 7.18 g
Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g
More about the ideal gas equation link is given below.
brainly.com/question/4147359
#SPJ1
Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L