Using a more concentrated HCl solution and Crushing the CaCO₃ into a fine powder makes the reaction to occur at a faster rate.
<u>Explanation:</u>
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(aq) + CO₂(g)
When calcium carbonate reacts with hydrochloric acid, it gives out carbon-dioxide in the form of bubbles and there is a formation of calcium chloride in aqueous medium.
The rate of the reaction can be increased by
- Using a more concentrated HCl solution
- Crushing the CaCO₃ into a fine powder
When concentrated acid is used instead of dilute acid then the reaction will occur at a faster rate.
When CaCO₃ is crushed into a fine powder then the surface area will increases thereby increasing the rate of the reaction.
The correct reaction equation is:

Answer:
b) 1 mole of water is produced for every mole of carbon dioxide produced.
Explanation: <u>CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.</u>
a) <u>22.4 L of
gas</u> is produced only when <u>
L of
</u> is reacted with 22.4 L of
. So it is wrong.
b) Since in the chemical equation the stoichiometric coefficient of
and
are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. <u>Therefore it is correct option.</u>
c)
molecules is equal 1 mole of
if produced then 3 moles of
is required, which is not given in the option. So it is wrong.
d) 54 g of water or 3 moles of
(<em>Molecular Weight of water is 18 g</em>) is produced when 3 moles of
is used but in this option only one mole of
is given. So it is wrong.
Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4

Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of
and
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:


Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M
Explanation:
The given data is as follows.
Energy of radiation absorbed by the electron in hydrogen atom = 
As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.
E = 
=
= 
or,
=
It is known that, 
= 
And, according to De-Broglie equation 
as, p = 
So, 
= 
Now, on squaring both the sides we get the following.
=
=

where, m = mass of electron
So, 
= 
=
J
Since, K.E = 
= 
= 
Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is
.