Answer :
(a) 0.0152
(b) 0.0000000778
(c) 0.000001
(d) 1600.1
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as 
889.9 is written as 
In this examples, 5000 and 889.9 are written in the standard notation and
and
are written in the scientific notation.
(a) 
The standard notation is, 0.0152
(b) 
The standard notation is, 0.0000000778
(c) 
The standard notation is, 0.000001
(d) 
The standard notation is, 1600.1
The answer is B particles combine into a large amount of engery.
You have to figure out a way to write the two unknown abundances in terms of one variable.
The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).
203(X) + 205(1-X) = 204.4
Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.
After that the higher percentage would be the most abundant.
203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6
x = 0.3
1-x = 0.7
Then the TI-205 would have the highest percentage and would be the most abundant.
Answer:
1,063 grams H₃PO₄
Explanation:
To find the mass of phosphoric acid (H₃PO₄), you should (1) convert molecules to moles (via Avogadro's number) and then (2) convert moles to grams (via molar mass from periodic table).
Molar Mass (H₃PO₄): 3(1.008 g/mol) + 30.974 g/mol + 4(15.998 g/mol)
Molar Mas (H₃PO₄): 97.99 g/mol
6.534 x 10²⁴ molecules H₃PO₄ 1 mole 97.99 g
--------------------------------------------- x ------------------------------------- x --------------
6.022 x 10²³ molecules 1 mole
= 1,063 grams H₃PO₄
Answer:
2.15 mg of uranium-238 decays
Explanation:
For decay of radioactive nuclide-

where N is amount of radioactive nuclide after t time,
is initial amount of radioactive nuclide and
is half life of radioactive nuclide
Here
,
and 
So,
so, N = 2.446 mg
mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg