Problem 2
You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.
Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.
Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.
Table
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
Rules
As the number of bonds INCREASES, the energy within the bond also INCREASES
As the number of bonds INCREASES, the distance of the bond DECREASES.
E = mc²
where E = energy produced
m = mass of the nucleus
C = speed of light
m = 9.106 x 10⁻³ x 1.67 x 10⁻²⁷ kg
C = 3 x 10⁸ m/s, thus C² = 9 x 10¹⁶
E = 1.37 x 10⁻¹² J
Response:
of 
can be found in 39.5 grams of .
Clarification:
Atomic weights: P= 31, F= 19,
The molar mass equals 1 atomic weight of P + 5 atomic weights of
F= 31+5 × 19
= 31+95
=126 g/mole
The number of moles in 39.5 gm of
equals 
= 
=0.3134 moles
1 mole of any substance encompasses
0.3131 moles comprises 0.3134
Thus, of can be found in 39.5 grams of .
Hello, in this situation, the chemical reaction occurring is as follows: Next, we will ascertain the limiting reactant by calculating the moles of magnesium oxide produced from 3.86 g of magnesium and 155 mL of oxygen using the given mole ratios of 2:1:2 and applying the ideal gas equation, demonstrating that oxygen is the limiting reactant because it generates the least magnesium oxide. Subsequently, we determine the mass of magnesium consumed solely by the oxygen.
