A(n) ________ microscope keeps an object in focus when the objective lens is changed.
1 answer:
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Answer:
m =8.81*10^{-6}grams
time t = 52.8 year
Explanation:
GIVEN DATA:
the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s
activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s
activity ,
= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2
= 8.877*10^{16}
Number of moles:
n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol
mass of the CO-60 is,
m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams
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time t = -[T1/2 / ln2]*ln[N/N0]
= - [5.3 years / ln2]*ln[1x10-6/1x10-3]
= 52.8 year
The mixing ratio is 6.
To find the answer, we have to know about the mixing ratio.
<h3>What is mixing ratio?</h3>- The mixing ratio must be calculated in a complex manner.
- A saturated vapor pressure (es) for values of air temperature and an actual vapor pressure (e) for values of dewpoint temperature must be determined in order to determine the mixing ratio.
- The air temperature and/or dewpoint temperature must first be converted to degrees Celsius (°C) before the vapor pressures can be calculated.
- The equation below can be used to determine the relative humidity (rh), as well as the actual mixing ratio and saturated mixing ratio,
where; w is the mixing ratio and w(s) is the saturation mixing ratio.
- In our question, it is given that,
- Thus, the mixing ratio will be,
Thus, we can conclude that, the mixing ratio is 6.
Learn more about mixing ratio here:
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