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.
Answer:
823.7g
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
Using the information given in this question as follows:
Q = 6,400 J
m = ?
c of soil = 0.840 J/g°C
∆T = 9.25°C
Using Q = mc∆T
m = Q ÷ c∆T
m = 6,400 ÷ (0.840 × 9.25)
m = 6400 ÷ 7.77
m = 823.7g
Answer: A golfer hitting a golf ball.
Explanation:
The atomic particles move more in this option than the others.
Oxidation state of Pb in PbO2 is +4.
Oxidation state of Pb in PbCl2 is +2.
Oxidation state of Pb in Pb2O is +1.
Oxidation state of Pb in Pb4O3 is +6/4.
Hence option A. PbO2 is correct.
Hope this helps, have a nice day!
<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
