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: That would be false because it is the contact between two layers representing a gap in the geologic record, usually from the erosion of the layers which would normally be expected to appear.
Explanation:
Have a good day
I hope this helps if not sorry :(
Stay motivated
1. Ca(HCO3)2
2.Ca(HCOO)2
3. Ca(OH)2
4.NaOH
5.KCI
6.MgSO4
7.PbO
8.HCl
9.HNO3
10.H2SO4
11.NH3
12.(NH4)3PO4
13.NaOH
:)
Answer:
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:
Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
![[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D2.0x10%5E%7B-3%7D%5Cfrac%7BmolCa%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B2molF%5E-%7D%7B1molCa%5E%7B2%2B%7D%7D%20%20%5C%5C)
Best regards.
If the.pressure exerted by a gas at [math]25^{\circ} \mathrm{C}[/math] in a volume of 0.044 L is 3.81 atm, how many moles of gas are present
