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nirvana33 [79]
3 years ago
7

You have just isolated a new radioactive element. If you can determine its half-life, you will win the Nobel Prize in physics. Y

ou purify a sample of 2 grams. One of your colleagues steals half of it, and six days later you find that 0.2 gram of the radioactive material is still left. What is the half-life
Chemistry
1 answer:
lbvjy [14]3 years ago
3 0

Answer: 2.58 days

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = age of sample  = 6 days

a = initial amount of the reactant  =  1 g

a - x = amount left after decay process   = 0.2 g

a) to find the rate constant

6=\frac{2.303}{k}\log\frac{1.0}{0.2}

k=\frac{2.303}{6}\log\frac{1.0}{0.2}

k=0.268days^{-1}

b) for completion of half life:  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.268}=2.58days

The half life is 2.58 days

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emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0
3 years ago
The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t
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<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
Potassium
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Answer:

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gtnhenbr [62]

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Conversion of minus 1 coulomb meter in debye
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Explanation:

Given:

Coulomb meter = -1 CM

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In debye

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