what?? please reword this
Answer:
1s2 2s2 2p6 3s2 3p6 4s2 3d5
Explanation:
According to the Aufbau principle, electrons are filled in orbitals in order of increasing energy. The energy of orbitals in the electronic configuration of manganese increases from left to right, hence 3d orbital is much greater in energy than a 3p orbital.
The arrangement of orbitals in order of increasing energy is shown in the answer above.
Answer is: line be long 3,011·10¹³ kilometers.
diametar of virus = 5·10⁻⁶ cm ÷ 100000 = 5·10⁻¹¹ km.
line lenght = 5·10⁻¹¹ km · 6,023·10²³.
line lenght = 3,011·10¹³ km.
Avogadro number = 6,023·10²³.
1 cm = 10⁻² m = 10⁻⁵ km.
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
It is option A 4s cause n=4, l=1 this is 4s orbital
