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.
1) To find the molar mass of C6H8O6, you must refer to the atomic weights of C, H, and O from the periodic table: C is 12; H is 1; O is 16 <span> (12x6)+(1x8)+(16x6)= 176g/mol
</span> <span> 176 g = 1 mol
0.5 g = x mol = 500 mg = 0.5 grams
Molar mass = mass ÷ moles
176 = 0.5 ÷ x
x = 2.84 x 10⁻³ mol
2) To calculate the total number of molecules in those </span> 2.84 x 10⁻³ mol, multiply the moles by <span> Avogadro's constant.
Number of molecules = Avogadro's constant x number of moles
Number of molecules = 6.022 x 10²³ x 2.84 x 10⁻³ </span> = 1.71 x 10²¹ molecules of vitamin C. <span>
</span>
Answer:
The molality is 1.15 m.
Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.
Calculate moles of H₂SO₄ from molarity:
C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles
Mass of solvent (water) based on density:
m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg
Therefore, molality is:
m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m
