Answer:
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
Explanation:
Answer:
[O2(g)][SO2(g)]^2/[SO3(g)]^2
Answer:
Q = 2640.96 J
Explanation:
Given data:
Mass of He gas = 10.7 g
Initial temperature = 22.1°C
Final temperature = 39.4°C
Heat absorbed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 39.4°C - 22.1°C
ΔT = 17.3°C
Q = 10.7 g× 14.267 J/g.°C × 17.3°C
Q = 2640.96 J