Answer:
2.765amu is the contribution of the X-19 isotope to the weighted average
Explanation:
The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:
X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21
The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:
19.00amu * 0.1455 =
2.765amu is the contribution of the X-19 isotope to the weighted average
The equilibrium constant is found by [product]/[reactant]
If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.
At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.
At equilibrium, for your reaction, it is predominantly reactants.
Answer:
The primary role of the carbonic-acid-bicarbonate buffer system is to neutralize the hydronium ions forming carbonic acid and water when any acidic substance enter the bloodstream.
Explanation:
hope it helps.
Answer: Option (3) is the correct answer.
Explanation:
When there is a negative charge on an atom then we add the charge with the number of electrons. Whereas when there is a positive charge on an atom then we subtract the charge from the number of electrons.
Atomic number of chlorine is 17. So, number of electrons present in 
is 17 + 1 = 18 electrons.
Atomic number of cobalt is 27. So, number of electrons present in 
is 27 - 4 = 23 electrons.
Atomic number of iron is 26. So, number of electrons present in 
is 26 - 2 = 24 electrons.
Atomic number of vanadium is 23. So, number of electrons present in V is 23 electrons.
Atomic number of scandium is 21. So, number of electrons present in 
is 21 + 2 = 23 electrons.
Thus, we can conclude that out of the given species, has the greatest number of electrons.
