Answer:
3.125% will remain after 125 days
Explanation:
Given data:
Half life of Th-234 = 25 days
Percent of thorium remain after 125 days = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 125 days / 25 days
Number of half lives = 5
At time zero =100%
At 1st half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At third half life = 25%/ 2 = 12.5%
At 4th half life = 12.5% /2 = 6.25%
At 5th half life = 6.25% /2 = 3.125%
Answer:
The mass percentage of bromine in the original compound is 81,12%
Explanation:
<u>Step 1: Calculate moles AgBr</u>
moles AgBr = mass AgBr / molar mass AgBr
= 0.8878 g / 187.77 g/mol
= 0.00472812 moles AgBr
⇒
Since 1 mol AgBr contains 1 mol Br-
Then the amount of moles Br- in the original sample must also have been 0.00472812 moles
<u>Step 2:</u> Calculating mass Br-
mass Br- = molar mass Br x moles Br-
= 79.904 g/mol x 0.00472812 mol
= 0.377796 g Br-
⇒
There were 0.377796 g Br- in the original sample
<u>Step 3:</u> Calculating mass percentage Br-
⇒mass percentage = actual mass Br- / total mass x 100%
% mass Br = 0.377796 g / 0.4657 g x 100 %
= 81.12%
Answer:
Depending on where ozone resides, it can protect or harm life on Earth. Most ozone resides in the stratosphere (a layer of the atmosphere between 10 and 40 km above us), where it acts as a shield to protect Earth's surface from the sun's harmful ultraviolet radiation.
Explanation:
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