Answer:
About 547 grams.
Explanation:
We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.
To do so, we can use the initial value and convert it to grams using the molar mass.
Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

Dimensional Analysis:

In conclusion, about 547 grams of copper (II) bicarbonate is produced.
Answer:
is larger
Explanation:
, where
is the acid dissociation constant.
For a monoprotic acid e.g. HA,
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the
value , lower will the the
In this mixture, at equilibrium,
will be constant.
of HF is grater than
of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So,
is larger
Answer:
Its because non-metals are unable to break the bond between the H and O ion and cannot reduce hydrogen by donating electrons
Answer:

Explanation:
Hello,
In this case, since the given 5-M concentration of magnesium chloride is expressed as:

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

Best regards.
Chemical Reactions and Moles of Reactants and Products
That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide.