Answer:
Explanation:
concepts, such as the internal energy of a system; heat or sensible heat, which are defined as types of energy transfer (as is work); or for the characteristic energy of a degree of freedom in a thermal system {\displaystyle kT}kT, where {\displaystyle T}T is temperature and {\displaystyle k}k is the Boltzmann constant.
Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as
Where 
is the quantum number
now
We can look at the given angular momentum in this form as
comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
Position of element in periodic table is depend on the electronic configuration of element.
Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.
Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.
Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table.
Answer:
C3 H6 O2
Explanation:
first divide their mass by their respective molar mass, we get:
30.4 moles of C
61.2 moles of H
20.25 moles of O
now divide everyone by the smallest one of them then we get
C= 1.5
H= 3
O= 1
since our answer of C is not near to any whole number so we will multiply all of them by 2
so,
C3 H6 O2 is our answer
Answer:
The value of the heat capacity of the Calorimeter 
= 54.4 
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water 
= 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q = 
ΔT + ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 + × 11.24
611.37 = × 11.24
= 54.4
This is the value of the heat capacity of the Calorimeter.
