Answer:
100 grams of C-14 decays to 25 grams in 11,460 years.
The C-14 isotope is only useful for dating fossils up to about 50,000 years old
If an ancient bone contains 6.25% of its original carbon, then the bone must be 22,920 years old.
Explanation:
We already know that the half life of C-14 is 5,730 years. After the first half life, we have 50 grams remaining. This takes 5,730 years. After the second half life (11,460 years now gone) we have 25 grams of C-14 left.
If a fossil material is older than 50,000 years an undetectable amount of 14C is left in the sample hence Carbon-14 is no longer suitable for dating the sample.
From;
0.693/5730 = 2.303/t log (No/0.0625No)
Where;
t = time taken and No = initial amount of C-14
0.693/5730= 2.77/t
t = 22,920 years
The removal of trees would most likely decrease the amount of carbon in the atmosphere.
Please note that it is useful to add the options provided with the question, in order to get a most accurate answer and have your question answered quicker.
Hope this helps!!
Hello there!
From my calculations the answers are:
D. Mass
B. Gravity
B. Gravity
Hope This Helps You!
Good Luck :)
Answer:
- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL
Explanation:
Hello,
In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:
- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.
Regards.
Answer:
The answer is B (25%)
Explanation:
I think the answer is this because it's the smallest value in the options, and if you choose 0% the answer would be wrong.
