We have gravity here on earth while in space there’s none
Answer:
Types of atomic orbitals present in the third principal energy are <u>s, p and d only .</u>
Explanation:
- <u>OPTION A-: s and p atomic orbitals -</u> these two orbitals are present in second principal energy level. Therefore , the option is incorrect.
- <u> OPTION B-: p and d only -</u> This option is wrong as there is no such principal level energy where , s atomic orbital is absent .
- <u>OPTION C-: s , p and d only -</u>these orbitals are present in<u> third principal energy level</u>. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .
- <u>OPTION D-: s , p, d and f only -</u>There is also a f sublevel at the <u>fourth and higher stages,</u> containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .
<u>Thus , the correct option is C (s , p and d only .)</u>
Pretty sure it is Block A:)
They have seven electrons in their valence shell, so halogens are very reactive.
Hope this helps! :)
According to this formula when:
ΔG = ΔH - TΔS = 0
∴ ΔS = ΔH/T
∴ ΔS = n*ΔHVap / Tvap
- when n is the number of moles = mass/molar mass
when the mass = 24.1 g
and the molar mass = 187.3764 g/mol
by substitution:
∴ n = 24.1 / 187.3764g/mol
= 0.129 moles
and ΔHvap is the molar enthalpy of vaporization is 27.49 kJ/mol
and Tvap is the temperature in Kelvin = 47.6 + 273 = 320.6 K
So by substitution, we will get the ΔS the change in entropy:
∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K
= 11 J/K
