Chemistry, 26.11.2019 17:31 eyebang

An element has three naturally occurring isotopes. use the information below to calculate the weighted average atomic mass of the element, showing both the setup and the final answer for the calculation. isotope atomic mass percent abundance i 39.96 u 99.59% ii 37.96 u 0.065% iii 35.97 u 0.336%

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Answer from: karankhatri1946

In order to calculate an average, we should sum all numbers and divide them by quantity.

Let’s work with qualifications first. Let’s say you got a 10 in 1 exam, then an 8 in 2 exams and a 4 in 2 exams. Your average will be:

= (10*1+8*2+4*2) / 5 = 6.8

If 6 is the minimum, you will pass.

There is another way to calculate this average: applying distributive property.

= 10*1/5+8*2/5+4*2/5 = 6.8

Remember you can convert the fractions into equivalent fractions: 1/5 = 20/100; 2/5 = 40/100

= 10*20/100+8*20/100+4*20/100 = 6.8

We actually don’t have the number of atoms of each mass… we have the percentage instead! So we need to learn this last method for atoms.

Let’s go back to our atoms problem:

99.59% of atoms have a mass of 39.96 u
0.065% of atoms have a mass of 37.96 u
0.336% of atoms have a mass of 35.97 u

So let’s put that in the formula:

Average mass = 39.96u*99.59/100 + 37.96u*0.065/100 + 35.97u*0.336/100

So what you have to know is that a percentage can be converted into a fraction, and you should work that fraction in order to find the average. We can make the calculus shorter putting 100 as the common denominator:

Average mass = (39.96u*99.59 + 37.96u*0.065 + 35.97u*0.336)/100

So actually we are taking the percentage as if it was the quantity, and 100 as if it was the total (the total of all percentages is always 100). Maybe we don’t have 100 atoms, but it will be the same proportion anyway, whatever number we have! And here it is the result:

Average mass = 39,94 u

Answer from: Quest

The answers is bad germs in their

Answer from: Quest

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.