(99.63/100 x 14.003) + (0.37/100 x 15)
13.9511889 + 0.0555 = 14.0066889amu
13.95 + 0.0555 = 14.0055 (same thing just rounded numbers)
(You can round that to 14amu)
Answer is 14 amu
Answer:
The answer to your question is 203.8 g
Explanation:
Data
Average atomic mass = ?
Centium-200
Centium-203
Centium-209
Equal amounts
Process
1.- Determine the abundance of each isotope.
If they are in equal amounts in nature, just divide 100 by 3
100/3 = 33.33 % or 0.333
2.- Determine the Average atomic mass
Average atomic mass = Atomic mass 1 x abundance + Atomic mass 2 x
abundance + Atomic mass 3 x abundance
- Substitution
Average atomic mass = 200 x 0.333 + 203 x 0.333 + 209 x 0.333
- Simplification
Average atomic mass = 66.6 + 67.6 + 69.6
Result
Average atomic mass = 203.8 g
The cornea is the outer clear, round structure that covers the Iris and the pupil. The Iris is the colored part of the eye.
Answer:
25 mM Tris HCl and 0.1% w/v SDS
Explanation:
A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:
(250 mM Tris HCl)/10 = 25 mM Tris HCl
(1.92 M glycine)/10 = 0.192 M glycine
(1% w/v SDS)/10 = 0.1% w/v SDS
Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.
