Answer:
104.969 amu.
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 107.977 amu
Abundance (A%) = 0.1620%
Isotope B:
Mass of B = 106.976 amu
Abundance (B%) = 1.568%
Isotope C:
Mass of C = 105.974 amu
Abundance (C%) = 47.14%
Isotope D:
Mass of D = 103.973 amu
Abundance (D%) = 51.13%
Average atomic mass =?
The average atomic mass of the element can be obtained as follow:
Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]
Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]
= 0.175 + 1.677 + 49.956 + 53.161
= 104.969 amu
Therefore, the average atomic mass of the element is 104.969 amu.
Answer:
The further an electron is from the nucleus. the greater its energy level.
Explanation:
When an electron is close to the nucleus, it is at as low an energy level as it can get.
We must put energy into an electron to pull it away from the attraction of a nucleus.
So, electrons that are further from the nucleus are at higher energy levels.
Answer:
81.04°C
Explanation:
Heat loss by water = Heat gained by Aluminum
Heat loss by water;
H = MCΔT
ΔT = 100 - T2
M = 580g
c = 4.2
H = 580 * 4.2 (100 - T2)
H = 243600 - 2436T2
Heat ganed by Aluminium
H = MCΔT
ΔT = T2 - 24
M = 900g
c = 0.9
H = 900 * 0.9 (T2 - 24)
H = 810 T2 - 19440
243600 - 2436T2 = 810 T2 - 19440
243600 + 19440 = 810 T2 + 2436T2
263040 = 3246 T2
T2 = 81.04°C
Assumption;
Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.
Answer:
it's natural selection
Explanation:
adaptation is when you adapt to some situation
