The answer is B ( you could have just looked it up online)
Answer:
c) 9.03 x 10^23
Explanation:
find the molar mass of Al
Al is 27.0 grams
Then use that, to find the number of moles in Aluminum.
Then use Avogadro's number which is 6.02 * 10^23
After that, write all of that down with dimensional analysis.
40.5 g * 1 mol/ 27.0 g of Al * 6.02 x 10^23 / 1mol
As your final answer, you will get 9.03 * 10^23 atoms with sig figs.
Hope it helped!
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
Answer: genus
Explanation:
Amoebas do not form a single taxonomic group; instead, they are found in every major lineage of eukaryotic organisms. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and animals.
