Molar mass of COH1206 is 1243.5857 g/mol
Considering; graphite; standard enthalpy = 0 and entropy = 5.740; diamond standard enthalpy = 1.897 and entropy = 2.38.
Using the equation Delta G = Delta H - Temperature (DeltaS)
Delta H = enthalpy sum of products - enthalpy sum of reactants
Which will be; 0 -1.897 = -1.897 kJ/Mol
Delta S is the entropy sum; given by
5.740 - 2.38 = 3.36 J/Mol
We can convert Delta S from Joules to kilo Joules by dividing by 1000
we get ; 0.00336 kJ/mol
We are given a temperature in kelvin which suits the calculations ((298 k)
Therefore; using the equation;
= -1.897 - (298 × 0.00336) = -2.90 kJ
Thus; the standard gibbs free energy will be; -2.9 kJ
Answer:
The magnesium reacted with the oxygen in the air.
Explanation:
For argument’s sake, let’s say that the mass of magnesium oxide was 3 g and that of the oxide was 5 g.
The reaction was
magnesium + oxygen ⟶ magnesium oxide
Mass: 3 g 5 g
Mass of oxygen = 5 g – 3 g = 2 g
The 3 g of magnesium must have combined with 2 g of oxygen to form 5 g of magnesium oxide.
According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
Answer:
Explanation:
As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.
Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>
So, you must just convert the given temperature, 32°F, to kelvin scale.
You can do that in two stages.
- First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80
- Second, convert 0°C to kelvin:
T (K) = T(°C) + 273.15 K= 273.15 K
Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K
Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.
