Let us check each statement one by one
a) Sb has a lower ionization energy but a higher electronegativity than I. : As per values given : Definitely Sb has lower ionization energy however the electronegativity of Sb is lower than that of iodine
b) Sb has a higher ionization energy but a lower electronegativity than I. FAlse:
Sb has lower ionization energy than I
c) Sb has a lower ionization energy and a lower electronegativity than I. True
d) Sb has a higher ionization energy and a higher electronegativity than I. False
Answer:
1st Blank: <em>1 Co</em>
2nd Blank:<em> 2 Na2S</em>
3rd Blank:<em> 4 Na</em>
4th Blank:<em> 1 CoS2</em>
Explanation:
<em>Trust me</em>
Delta T= T final - T initial
Tfinal= -101.1 °C
Tinitial= -0.5 °C
•Delta T = -101.1°C - (-0.5°C)
=100.6°C
Kelvin= °C + 273
= -100.6 + 273
= 172.4 Kelvin
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
