14 since K has 1 valence but there’s two so 2 valence for k and oxygen has 6 but there’s two so 12
Answer:
157.64 L
Explanation:
We'll begin by converting 30 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 30 °C
T(K) = 30 °C + 273
T (K) = 303 K
Next, we shall convert 600 mmHg to atm. This can be obtained as follow:
760 mmHg = 1 atm
Therefore,
600 mmHg = 600 mmHg × 1 atm / 760 mmHg
600 mmHg = 0.789 atm
Finally, we shall determine the volume of the gas. This can be obtained as follow:
Number of mole (n) = 5 moles
Temperature (T) = 303 K
Pressure (P) = 0.789 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
0.789 × V = 5 × 0.0821 × 303
0.789 × V = 124.3815
Divide both side by 0.789
V = 124.3815 / 0.789
V = 157.64 L
Therefore, the volume of the gas is 157.64 L
Answer:
Keqq = 310
Note: Some parts of the question were missing. The missing values are used in the explanation below.
Explanation:
<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>
Using the formula ΔG° = -RT㏑Keq
㏑Keq = ΔG°/(-RT)
where ΔG° = ΔH° - TΔS°
㏑Keq = ΔH° - TΔS°/(-RT)
㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)
㏑Keq = -14221.2/-2478.968175
㏑Keq = 5.73674166
Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶
Keq = 310.05
One is a mixture and the other is a compound