atomic mass=percentage of isotope a * mass of isotope a + percentage of isotope b * mass of isotope b+...+percentage of isotope n * mass of isotope n.
Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%
mass of isotope₂=270.9 u
percentage of isotope₂=9.7%
Therefore:
atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u
Answer: the mass atomic of this element would be 268.1 u
<span>There is five main area of study in Chemistry, these are:
Analytical, this focusses on experimental equipment and methods used in chemistry (e.g., NMR, Spectroscopic methods, etc.)
Biochemistry - focuses on the chemistry of compounds and processes in living things (e.g., amino acids, proteins, DNA, cellular respiration, Krebs cycle, etc.)
Organic - focuses on the chemistry on most carbon-based molecules found in living things (e.g., hydrocarbons, alcohols, carbolic acids. Amines, ester, etc.)
Inorganic - (focuses on all elements other than carbon (e.g., fluorine, silicon, xenon, etc.)
Physical - focuses on the basic structure and energetic son atoms and molecules (e.g., subatomic structure, is nice and covalent bonding, thermodynamics, reactions, etc.)</span>
Answer:
the tertiary animals the bigger animals that feed on smaller fish and crustaceans. These include predators like sharks, barracuda and tuna snapper. these fish that are commercially fished at unsustainable levels.
Explanation:
Answer:
D. Propanol
Explanation:
C3H7OH the presence of alcohol functional group makes it propanol
Explanation:
The given data is as follows.
= 0.483, ![[Co^{3+}]](https://tex.z-dn.net/?f=%5BCo%5E%7B3%2B%7D%5D)
= 0.173 M,
![[Co^{2+}]](https://tex.z-dn.net/?f=%5BCo%5E%7B2%2B%7D%5D)
= 0.433 M, ![[Cl^{-}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D)
= 0.306 M,
= 9.0 atm
According to the ideal gas equation, PV = nRT
or, P = 
Also, we know that
Density = 
So, P = MRT
and, M = 
= 
= 
= 0.368 mol/L
Now, we will calculate the cell potential as follows.
E = ![E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCo%5E%7B2%2B%7D%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%7D%7B%5BCo%5E%7B3%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D%7D)
= 
= 
= 
= 0.483 - 0.0185
= 0.4645 V
Thus, we can conclude that the cell potential of given cell at 
is 0.4645 V.
