The radioactive nuclide Ta-170 undergoes first-order decay with a half-life of 6.76 min. If a quantity of Ta-170 is produced, what fraction remains after 75.1 seconds

0.88

Explanation:

Step 1 Obtain the decay constant(λ) from

t1/2 = 0.693/λ

λ = 0.693/t1/2

t1/2 = half life = 6.76 min = 405.6 secs

λ = 0.693/405.6 secs

λ = 0.0017 s-1

Step 2 Obtain the fraction left at a time t

From

N= Noe^-λt

Where;

N= mass of radioactive material left after a time t

No = mass of radioactive material originally present

λ = decay constant = 0.0017 s-1

t= time taken = 75.1 secs

Hence;

N/No = e^-0.0017 * 75.1

N/No = 0.88

Note; N/No is the required fraction left after 75.1 seconds

a is the answer

explanation:

which statement about trans-fatty acids is false?

a. the conversion of cis-fatty acids to trans-fatty acids is inhibited by the presence of antioxidants.

hope this

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.