







There are two critical points, at which we have


22.7
is the anwser
because irrational is repeating
Answer:
4. 80°
5. 11
6. 9
Step-by-step explanation:
4. 80 is the lowest number on the chart
5. add temperatures between 86 and 89
6. add temperatures between 86 and 90
Answer:
average atomic mass = 207.2172085 amu
Explanation:
To get the average atomic mass of an element using the abundance of its isotopes, all you have to do is multiply each isotope by its percentage of abundance and then sum up all the products.
For the question, we have:
203.97304 amu has an abundance of <span>1.390% (0.0139)
</span>205.97447 amu has an abundance of 24.11% (0.2411)
206.97590 amu has an abundance of 22.09% (0.2209)
207.97665 amu has an abundance of 52.41 % (0.5241)
The average atomic mass can be calculated as follows:
average atomic mass = 203.97304(0.0139) + 205.97447(0.2411)
+ 206.97590(0.2209) + 207.97665(0.5241)
average atomic mass = 207.2172085 amu
Hope this helps :)