I believe it would be D. Electromagnet. It's been a while since I've done this stuff, tho. Hope this helps!!!! :)
B) with single bonds..... Hope it helps, Have a nice day :)
Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>
The answer is 2H2 + O2----> 2H2O