Answer:
After 90 minutes 5.15625 g amount left from total of 165 g.
Explanation:
Given data:
Technetium-104 half life = 18.0 min
Total amount of sample = 165 g
Amount left after 90.0 min = ?
Solution:
First of all we will calculate the number of half lives passes.
Number of half lives = T elapsed / Half life
Number of half lives = 90 min /18.0 min
Number of half lives = 5
Amount left:
At time 0 = 165 g
At first half life = 165 g/ 2= 82.5 g
At 2nd half life = 82.5 g/2 = 41.2 g
At 3rd half life = 41.2 g/ 2 = 20.625 g
At 4th half life = 20.625 g/ 2 = 10.3125 g
At 5th half life = 10.3125 g /2 = 5.15625 g
Thus, after 90 minutes 5.15625 g amount left from total of 165 g.
An original sample of K-40 has a mass of 25.00 grams.
After 3.9 m 109 years, 3.125 grams of the original sample remains unchanged.
What is the half-life of K-40?
First step is to determine the remaining decimal amount.
3.125 grams /25.00 grams = 0.125
Second step is to determine the number of half lives.
(1/2)^n = 0.125
N log (1/2) = log 0.125
N = 3 years
Yea same I tried so hard I think you need to get a expert or look this up on a website or something
D. For plato users
(I've had the exact same one)
The standard cell potential of a cell made of theoretical metals ma/ma2+ and mb/mb2+ is to be determined. We are given their reduction potentials:
-0.19 v for A
-0.85 v for B
To solve for the standard cell potential, examine the reaction and
the standard cell potential for this case is (-0.19- (-0.85)) v = -0.66 v<span />
