X

H has a positive 1 charge. This means that having 3H = +3<span>. This is a neutral compound so x= -3 because X+3H= 0
Y</span>

is also neutral so 2X+Y= 0
we know X=-3 So, 2(-3)+Y=0
-6+y=0
Y=+6 charge
Answer: The valency of X is -3.
The valency of Y is 6
Perchloric acid (HCO₄) is one of strongest inorganic acids. Perchloric acid precipitation is used to removes most of the protein present in the
sample and stabilize many of the small molecule analytes. It can use also to precipitation glycogen, ATP, glutathione, antioxidants.
Perchloric acid donates protons to macromolecules and lower their solubility in water. Also perchloric acid is strong oxidizing agent and change oxidation number of elements.
The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.
The alkali earth metal beryllium (
) belongs to the second group of the periodic table. The ground state electronic configuration of
is:
From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (
).
After losing all valence electrons that is 2 electrons from
orbital. The electronic configuration will be:

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as
.
Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.
If a piece of charcoal is taken and set on fire it will evolve CO2 and heat. Charcoal is solid but carbon dioxide is gaseous. This is an example of change of state and evolve of heat during chemical reaction.
Answer:
Explanation:
When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.
When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.
2H2O2 (aq) → O2(g) + 2H2O (l) ------------- first order kinetics reaction.
Catalysts are KI, FeCl3 only, KCl is not a catalyst. Order: KI < MnO2 < Pb < FeCl3.
H2O2 + I– -> IO– + H2O (Step 1)
H2O2 + IO– -> I– + H2O + O2 (Step 2)
It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.
02(g)2Fe? (aq) + 2 H(a) 2 H 2 Fe3 (aq) H2O2(aq) + 2 Fe,Taq) H02(aq) 2 Fe (aq) 2 H (aq)