Dangduted bot depende Horthe following pas-phase reaction is Bond | D (EJ/mol) A
ID: 698334 • Letter: D
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Dangduted bot depende Horthe following pas-phase reaction is Bond | D (EJ/mol) AJ 2012 2) or the ions below, only 8) -156 0-57 D) -291 Lhas a noble gas electron configuration. D) S3- E) 291 C) 02+ E) K- 3) A 285 mil. round-bottom flask is weighed and found to have a mass of 114.85 g. A few milliliters of 3) an easily vaporized liquid are added to the flask and the filask is immersed in a boiling water bath. All of the liquid vaporizes at the boiling temperature of water, filling the flask with vapor. When all of the liquid has vaporized, the flask is removed from the bath, cooled, dried, and reweighed. The new mass of the task and the condensed vapor is 115.23 g. Which of the following compounds could the liquid be? (Assume the ambient pressure is 1 atm.) ) 25 B) C4H10 c) 4 D) C2H6 E) C3H7OHExplanation / Answer
At the left you are breaking double bond C =C with an energy of 614 KJ mol
and you break an H Br bond with an energy of 366 KJ / mol
enthalpy of bonds broken = 614 + 366 = 980 KJ / mol
At the right you formed a C - C single bond 348
a C H bond 413 and
C Br bond 276
Bonds formed = 348 + 413 + 276 = 1037
Enthaly is = bonds broken - bonds formed = 980 - 1037 = -57 KJ / mol
2. The answer is Cl-, this is because in the periodic chart is very close to the Ar if you add just 1 electron the electron configuration is the same that Ar , the other ones are close to a noble gas but does not match any
3. You have a vapor with a mass of 115.23 - 114.85 = 0.38 grams of vapor, apply ideal gas equation
PV = n R T, p is pressure, V is volume, n is moles, R is gas constant and T is temperature
so
1* 0.255 Liters = n * 0.082 * (100 + 273), p is given in the statement 1 atm, the volume is the flask volume, temperature is 100 C but change it to Kelvin by adding 273 so
n = (0.255 ) / (0.082* 373) = 0.00833 moles
moles = mass / molar mass, lets change the equation to find molar mass
molar mass = mass / moles = 0.38 / 0.00833 = 45.61 g/gmol , now we only need to match this value with the molar masses of the compounds given
C2H5OH = 46.07 g/gmol
C4H10 is 58.12
C4H9OH is 74.12 g/gmol
C2H6 is 30
C3H7OH is 60
So the compound is very likely to be Ethanol C2H5OH
The next question are hard to read please post them differently with a better resolution
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Q 12. Valence electrons for CH3CH2Cl
C has 4 valence electrons
H has 1
Cl has 7 so, we have 2 carbons 5 hydrogens and 1 cl so electrons are = 2*4 + 1*5 + 1*7 = 8 + 5 + 7 = 20 valence electrons
Q 13
For polarity you have to look for the electronegativity value for each compound and substract the values, the bigger the difference the bigger the polarity, in the periodic chart electronegativity increases as you go from the left to the right and from the bottom to the top , so the first thing to see is how close are the elements in the periodic chart, from that point of view the compounds that must have a bigger polarity because their forming elements are separated in the periodic chart are Mg - F and Be - F
Mg and Be are very close, if you take a look for the particular electronagativity value you see that Mg has an a value of 1.3 and Be 1.6 , F has a value of 4 so the biggest difference comes from the Mg - F, so Mg - F must be the last one (most polar) Be - F the compound before that so answer is
O - F, N - F, Be - F and Mg - F , (oxygen and F are very close so the polarity is small)
Q 14
Ideal gas equation is PV = n RT
, n is mass / molar mass
PV = mass / mm R T
P = mass / (volume * molar mass ) * R T, mass / volume = density
P = density * R * T / molar mass
density = P * molar mass / R T, molar mass of kripton is 84, P is 1.21 atm and T is 50C from the statement
density kripton = 84 * 1.21 / (0.082 * 323) = 3.83 g/L
Van der waals equation recognizes....
Answer is E, all above are true
PF3 is trigonal pyramidal and is considered a polar molecule
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