Gold is the best pure substance. The answer is gold
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:
3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane
Both of these are correct. This is an alkane, because it has all single bonds.
B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:
4-methyl-2-pentyne
This is an alkene, because of the double bond.
C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).
4,6-dimethyl-2-octene
This is an alkene, because of the double bond.
D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:
1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene
Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.
E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:
3,5,7-trimethyl-5-propylnonane
This is an alkane, due to the single bonds.
Hope this helps!
Answer:
0.712 mol of NO₂ are formed .
Explanation:
For the reaction , given in the question ,
2 N₂O₅ ( g ) → 4 NO₂ ( g ) + O₂ ( g )
From the above balanced reaction ,
2 mol of N₂O₅ reacts to give 4 mol of NO₂
Applying unitary method ,
1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂
From the question , 0.356 mol of N₂O₅ are reacted ,
<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>
Since ,
1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂
0.356 mol of N₂O₅ reacts to give 4 / 2 * 0.356 mol of NO₂
Calculating ,
0.712 mol of NO₂ are formed .
Answer:
At the equivalence point, equal amounts of H+ and OH– ions will combine to form H2O, resulting in a pH of 7.0 (neutral). The pH at the equivalence point for this titration will always be 7.0, note that this is true only for titrations of strong acid with strong base.
Explanation:
