Explanation:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
The atomic mass of isotope (II) of vanadium is 49.944 amu.
Answer:
A --- (E)-oct-2-en-1-o1
B ----(E)-oct-2-enal
Explanation:
See the attached file for the structure.
To help, I drew a diagram. This represents an ionic bond between Na and Cl. Na is giving his single electron to Cl, which is indicated by the arrow, to make Cl full with 8 electrons.
Carbonated water would be what you are looking for :v)
Answer: 2.5°C
Explanation:
Initial volume V1 = 5.38 liters
Initial temperature T1 = 36.0°C
Convert temperature in Celsius to Kelvin
(32°C + 273= 305K)
Final temperature T2 = ?
Final volume V2 = 4.68 liters
According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.
Thus, Charles' Law is expressed as: V1/T1 = V2/T2
5.38/305 = 4.86/T2
To get the value of T2, cross multiply
5.38 x T2 = 4.86 x 305
5.38T2 = 1482.3
Divide both sides by 5.38
5.38T2/5.38 = 1482.3/5.38
T2 = 275.5K
Convert 275.5K to Celsius
(275.5K - 273K = 2.5°C)
Thus, the final temperature is 2.5°C
