Answer:
2–butyne.
Explanation:
To name the compound given above, we must determine the following:
1. Determine the functional group of the compound.
2. Determine the longest continuous carbon chain. This gives the parent name of the compound.
3. Locate the position of the functional group by giving it the lowest possible count.
4. Combine the above to obtain the name.
Thus, we shall name the compound as follow:
1. The compound contains triple bond (C≡C). Therefore, the compound is an alkyne.
2. The longest chain is carbon 4. Thus the parent is butyne.
3. The triple bond (C≡C) is located at carbon 2 when we count from either side.
4. The name of the compound is:
2–butyne
Molecular weight of oxygen is 15.998g therefor 1.5(2*15.998)=47.994g
Answer:
0.34 M
Explanation:
I assume that the compound is PbCl2.
One mole of PbCl2 contains one mol of Pb+2 and 2 moles of Cl-
Molarity (M)= moles (n) /Volume (V)
Moles Pb2+ = M x V = 0.17 V
Moles Cl- = moles Pb2+ x (2 moles Cl-/1 mole Pb2+) = 0.17 V x 2 = 0.34 V
M Cl- = moles Cl-/V = 0.34V/V = 0.34 M
When the overall charge is equal to zero
