Answer: a) 3.85 days
b) 10.54 days
Explanation:-
Expression for rate law for first order kinetics is given by:
![t=\frac{2.303}{k}\log\frac{a}{a-x}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7Bk%7D%5Clog%5Cfrac%7Ba%7D%7Ba-x%7D)
where,
k = rate constant = ?
t = time taken for decomposition = 3 days
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = ![\frac{58}{100}\times 100=58g](https://tex.z-dn.net/?f=%5Cfrac%7B58%7D%7B100%7D%5Ctimes%20100%3D58g)
First we have to calculate the rate constant, we use the formula :
Now put all the given values in above equation, we get
![k=\frac{2.303}{3}\log\frac{100}{58}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7B3%7D%5Clog%5Cfrac%7B100%7D%7B58%7D)
![k=0.18days^{-1}](https://tex.z-dn.net/?f=k%3D0.18days%5E%7B-1%7D)
a) Half-life of radon-222:
![t_{\frac{1}{2}}=\frac{0.693}{k}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B0.693%7D%7Bk%7D)
![t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B0.693%7D%7B0.18%7D%3D3.85days)
Thus half-life of radon-222 is 3.85 days.
b) Time taken for the sample to decay to 15% of its original amount:
where,
k = rate constant = ![0.18days^{-1}](https://tex.z-dn.net/?f=0.18days%5E%7B-1%7D)
t = time taken for decomposition = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = ![\frac{15}{100}\times 100=15g](https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B100%7D%5Ctimes%20100%3D15g)
![t=\frac{2.303}{0.18}\log\frac{100}{15}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7B0.18%7D%5Clog%5Cfrac%7B100%7D%7B15%7D)
![t=10.54days](https://tex.z-dn.net/?f=t%3D10.54days)
Thus it will take 10.54 days for the sample to decay to 15% of its original amount.
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
Complete Question
Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light
Answer:
The diameter is
Explanation:
From the question we are told that
The best resolution is ![\theta = 0.3 \ arcsecond](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%200.3%20%5C%20%20arcsecond)
The wavelength is ![\lambda = 700 \ nm = 700 *10^{-9 } \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20700%20%5C%20%20nm%20%3D%20%20700%20%2A10%5E%7B-9%20%7D%20%5C%20%20m)
Generally the
1 arcminute = > 60 arcseconds
=> x arcminute => 0.3 arcsecond
So
![x = \frac{0.3}{60 }](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B0.3%7D%7B60%20%7D)
=> ![x = 0.005 \ arcminutes](https://tex.z-dn.net/?f=x%20%3D%200.005%20%5C%20%20arcminutes)
Now
60 arcminutes => 1 degree
0.005 arcminutes = > z degrees
=> ![z = \frac{0.005}{60 }](https://tex.z-dn.net/?f=z%20%3D%20%20%5Cfrac%7B0.005%7D%7B60%20%7D)
=> ![z = 8.333 *10^{-5} \ degree](https://tex.z-dn.net/?f=z%20%3D%20%208.333%20%2A10%5E%7B-5%7D%20%20%5C%20degree)
Converting to radian
![\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20z%20%3D%20%208.333%20%2A10%5E%7B-5%7D%20%20%2A%200.01745%20%3D%201.454%20%2A10%5E%7B-6%7D%20%5C%20%20radian)
Generally the resolution is mathematically represented as
![\theta = \frac{1.22 * \lambda }{ D}](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20%5Cfrac%7B1.22%20%2A%20%20%5Clambda%20%20%7D%7B%20D%7D)
=> ![D = \frac{1.22 * \lambda }{\theta }](https://tex.z-dn.net/?f=D%20%3D%20%20%5Cfrac%7B1.22%20%2A%20%5Clambda%20%7D%7B%5Ctheta%20%7D)
=>
=>
To what i see, the answer is....
C.
Answer:
160000000 kg.
Explanation:
p=mv
p=1.6x10^9
v=10m/s
rearrange and substitute:
(1.6x10^9)=m(10)
m=(1.6x10^9)/10
m= 1.6x10^8 kg.