Answer:
6 moles
Explanation:
From the equation you can see that twice as many HCL moles are used as H2 produced
2 x 3 = 6 moles of HCL required
Answer:
Wavelength of the first four spectral line:




Explanation:
Lman series when electron from higher shell number and jumps to 1st shell then all the possible lines is called as lyman series.
to calculate wavelength of first four spectral line:
For hydrogen Z=1;
by using rydberg equation
![\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3DRZ%5E2%5B%5Cfrac%7B1%7D%7Bn_1%5E2%7D%20-%5Cfrac%7B1%7D%7Bn_2%5E2%7D%20%5D)
1. n=2 to n=1
![\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3DRZ%5E2%5B%5Cfrac%7B1%7D%7Bn_1%5E2%7D%20-%5Cfrac%7B1%7D%7Bn_2%5E2%7D%20%5D)
=rydberg constant


2. n=3 to n=1


3. n=4 to n=1


4. n=5 to n=1


Mv^2 = const
<span>m ~ 1/v^2 </span>
<span>(4.5/10.1)^2 = 0.198 </span>
<span>20.2 * 0.198 = 4.0 g/mol (Helium)</span>
Answer : The value of
for
is
.
Solution : Given,
Solubility of
in water = 
The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.
The equilibrium equation is,

Initially - 0 0
At equilibrium - s s
The Solubility product will be equal to,
![K_{sp}=[Ba^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)

![[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5BCO%5E%7B2-%7D_3%5D%3Ds%3D4.4%5Ctimes%2010%5E%7B-5%7Dmole%2FL)
Now put all the given values in this expression, we get the value of solubility constant.

Therefore, the value of
for
is
.
Answer: A
Explanation: Protons and neutrons form the nucleus of the atom, with electrons orbiting it.