Answer:
a) a0 was 46.2 grams
b) It will take 259 years
c) The fossil is 1845 years old
Explanation:
<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>
A = A0 * (1/2)^(t/h)
⇒ with A = the final amount = 46.2 grams
⇒ A0 = the original amount
⇒ t = time = 8 hours
⇒ h = half-life time = 3.2 hours
46.2 = Ao*(1/2)^(8/3.2)
Ao = 261.35 grams
<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>
t = (ln(0.66))-0.693) * 432 = 259 years
It will take 259 years
<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>
<em />
t = (ln(0.80))-0.693) * 5730 = 1845
The fossil is 1845 years old
Answer:
Your coefficients (the numbers in front of the molecule) will be the following from left to right.
1. <u>1 - 2 - 1 - 2</u>
2. <u>2 - 1 - 2 - 2 - 1</u>
3. <u>2 - 4 - 1</u>
4. <u>2 - 4 - 3</u>
5. <u>2 - 2 - 2 - 1</u>
6. <u>1 - 1 - 1</u>
7. <u>2 - 1 - 2</u>
8. <u>3 - 1 - 2 - 3</u>
9. <u>3 - 1 - 2 - 3</u>
10. <u>2 - 1 - 1 - 1</u>
Explanation:
To balance this equations first count how many times an element is on each side and then see what needs to be changed in order to balance them.
Answer:
Both Options C and D are appropriate.
But I'd go with Option D since "Direct Air Capture" would eventually lead to "Ground Injection"
OPTION D.
Answer: 1
Explanation: Lithium has a single electron in the second principal energy level, and so we say that lithium has one valence electron.
The 8 hour shift is from 7.00 am to 3.00 pm.<span>
<span>At the beginning, from 7.00 am to 9.30am (2.5 hrs.),<span> the client has normal saline which </span></span>is
infusing at 125 ml/hr.
Hence amount of normal saline which was injected= 2.5 hrs. x 125 ml/hr.
= 312.5 ml
from 9.30am to 3.00 pm (5.5 hrs) lactated ringer solution, which is to infuse at 100 ml/hr was
given.
Hence amount of lactated ringer solution given = 5.5 hrs x 100 ml/hr
= 550 ml
Total amount of IV solution given = normal saline
+ lactated ringer solution
= 312.5 ml + 550
ml
= 862.5 ml
<span>
= 863 ml</span></span>
