Can still be used to do work
I think this is what you're after:
Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1]
Convert to J and divide by the Avogadro Const to give E in J per photon
E = 375700/6.022×10^23 = 6.239×10^-19 J
Plank relationship E = h×ν E in J ν = frequency (Hz s-1)
Planck constant h = 6.626×10^-34 J s
6.239×10^-19 = (6.626×10^-34)ν
ν = 9.42×10^14 s^-1 (Hz)
IP are usually given in ev Cs 3.894 eV
<span>E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon </span>
<span>The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.</span>
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!
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
