Answer : The molar mass of the unknown gas will be 79.7 g/mol
Explanation : To solve this question we can use graham's law;
Now we can use nitrogen as the gas number 2, which travels faster than gas 1;
So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1
We can compare the rates of both the gases;
So here, Rate of gas 2 / Rate of gas 1 = 
Now, 1.687 = square root [
]
When we square both the sides we get;
2.845 = (molar mass 1) / (28.01 g/mol N2)
On rearranging, we get,
2.845 X (28.01 g/mol N2) = Molar mass 1
So the molar mass of unknown gas will be = 79.7 g/mol
Answer: A. The reaction takes place in one step.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
![Rate=k[A]^a[B]^b](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Ea%5BB%5D%5Eb)
k= rate constant
a= order with respect to A
b = order with respect to B
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
From,
RAM=element×its relative abudance/total abudance
=((107×13)+(12×109))/25
The answer is=107.96
