Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%
Explanation: Given that the average atomic mass(M) of magnesium
= 24.3050amu
Mass of first isotope (M1) = 23.9850amu
Mass of middle isotope (M2)=24.9858amu
Mass of last isotope(M3)= 25.9826amu
Total abundance = 1
Abundance of middle isotope = 0.10
Let abundance of first and last isotope be x and y respectively.
x+0.10+y =1
x = 0.90-y
M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope
24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y
Substitute x= 0.90-y
Then
y = 0.11
Since y=0.11, then
x= 0.90-0.11
x=0.79
Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%
Answer:
May be the instrument is incorrect or may be error in it.
Explanation:
The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.
Answer:
the mass of the air in the room is 4.96512 kg ( in 0°C)
Answer: Mass Of CFC that needs to evaporate for the freezing of water = 328.24 g
Explanation: Heat gained by the CFC = Heat lost by water
Heat lost by water = Heat required to take water's temperature to 0°c + Heat required to freeze water at 0°c
Heat required to take water's temperature from 33°c to 0°c = mCΔT
m = 201g, C = 4.18 J/(gK), ΔT = 33
mCΔT = 201 × 4.18 × 33 = 27725.94 J
Heat required to freeze water at 0°c = mL
m = 201g, L = 334 J/g
mL = 201 × 334 = 67134 J
Heat gained by CFC to vaporize = mH = 27725.94 + 67134 = 94859.94 J
H = 289 J/g, m = ?
m × 289 = 94859.9
m = 328.24 g
QED!!
