Answer:
There was a change in temperature on the surface you were walking on. It is possible that "you" were walking on cool, wet sand and when your parents said it was time to leave, you ran barefoot towards the parking lot (away from the sand) and got a burning coal sensation on your feet.
Explanation:
H20 can be a molecule or a compound because it is made up of Hydrogen(H)
and oxygen (O)
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
3 covalent bonds (there are 2 electrons in the first orbital and 5 in the second. You still have room for three more)
Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
