Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
The number of potassium atom that are in 0.25 moles potassium carbonate is calculated as follows
by use of Avogadro contant
1 mole= 6.02 x10^23 atoms
what about 0.25 moles,
by close multiplication
{0.250 moles x 6.02 x10^23} / 1 mole = 1.505 x10^23 atoms
Fluoride is an anion of Fluorine
What this means is that the two have the same number of protons (9), but Fluoride has 10 electrons compared to Fluorine's 9.
So the answers are:
Protons - 9
Neutrons - 9
Electrons - 10
Atomic Number - 9
Atomic Mass - 19 g/mol
- 407.4 kJ of heat is released.
<u>Explanation:</u>
We have to write the balanced equation as,
2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O
Here 2 moles of ethane reacts in this reaction.
Now we have to find out the amount of ethane reacted using its given mass and molar mass as,
2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆
Heat released = ΔH × given mass / 60.14
= - 1560. 7 kj ×15.7 g / 60. 14 g = -407. 4 kJ
