"The other halogens are not as electronegative and so other hydrogen halides cannot form hydrogen bonds between molecules. Only London Forces are formed. - Therefore more energy is required to break the intermolecular forces in HF than the other hydrogen halides and so it has a higher boiling point."
not a hack link, just stating where i got your answer from! -
https://www.mytutor.co.uk/answers/17558/A-Level/Chemistry/Explain-the-unusually-high-boiling-point-of-HF/
Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!
http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm
<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V
<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)
</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)
<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V
V= - 0.777V (no spontaneous)
<span>0.310 moles
First, look up the atomic weights of the elements involved.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight sulfur = 32.065
Molar mass (C3H5)2S = 6 * 12.0107 + 10 * 1.00794 + 32.065
= 114.2086 g/mol
Moles (C3H5)2S = 35.4 g / 114.2086 g/mol = 0.309959145 mol
Since there's just one sulfur atom per (C3H5)2S molecule, the number of moles of sulfur will match the number of moles of (C3H5)2S which is 0.310 when rounded to 3 significant digits.</span>
₁₇Cl 1s²2s²2p⁶3s²3p⁵, 1 unpaired electron in 3pz orbital.
