<h3>
Answer:</h3>
Chlorine gas (Cl₂)
<h3>
Explanation:</h3>
- According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
- Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
- Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.
In this case we are given gases, CI₂
, H₂,He and Ne.
- We are required to identify the gas that will diffuse at the slowest rate.
- In other words we are required to determine the heaviest gas.
Looking at the molar mass of the gases given;
Cl₂- 70.91 g/mol
H₂- 2.02 g/mol
He - 4.00 g/mol
Ne- 20.18 g/mol
Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
Answer:
Lathanum .
Atomic number = 57
Symbol = La
Atomic weight = 138.9
No of energy orbitals = 6
Electronic configuration
![[Xe]6s^25d^1](https://tex.z-dn.net/?f=%5BXe%5D6s%5E25d%5E1)
Answer: 72.4 kJ/mol
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_3H_8%7D%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=-2220.1%3D%5B%283%5Ctimes%20-393.5%29%2B%284%5Ctimes%20-241.8%29%5D-%5B%285%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)

Therefore, the heat of formation of propane is 72.4 kJ/mol
Answer: 281 hours
Explanation:-
1 electron carry charge=
1 mole of electrons contain=
electrons
Thus 1 mole of electrons carry charge=

of electricity deposits 1 mole or 63.5 g of copper
0.0635 kg of copper is deposited by 193000 Coloumb
11.5 kg of copper is deposited by=
Coloumb

where Q= quantity of electricity in coloumbs = 34952756 C
I = current in amperes = 34.5 A
t= time in seconds = ?


Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.