The answer should be 2KI(aq)+Hg(NO3)2(aq) = 2KNO3(aq) + HgI2(s). This is a double replacement reaction, the valence of elements will not change. The HgI2 is not soluble in water.
When Use a beaker you measure in ml.
I don’t have a picture but I can describe it to you.
The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.
The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.
Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.
Hope this helps!
Answer:
2.5
Explanation:
From the question given above, the following data were obtained:
Molarity of NaOH = 3.0x10¯³ M
pOH =?
Next, we shall determine the concentration of the hydroxide ion in the solution. This can be obtained as follow:
NaOH (aq) —> Na⁺ (aq) + OH¯ (aq)
From the balanced equation above,
1 mole of NaOH produced 1 mole of OH¯.
Therefore, 3.0x10¯³ M NaOH will also produce 3.0x10¯³ M OH¯.
Finally, we shall determine the pOH of the solution. This can be obtained as illustrated below:
Concentration of hydroxide ion [OH¯] = 3.0x10¯³ M
pOH =?
pOH = – Log [OH¯]
pOH = – Log 3.0x10¯³
pOH = 2.5
Thus, the pOH of the solution is 2.5
Answer:
C
Explanation:
Na has an electron configuration of 2,8,1
it forms an ion by losing 1 electron thus gaining a 1+ charge
