1- Molar solutions: based on number of moles of chemical in 1 litre of solution
2- Weight % solution: the weight of chemical divided by the total weight of the solution (chemical + water) and multiplied by 100.
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
Answer:
See below
Explanation:Plot the known concentrations and adsorbance data. Draw a best fit line through thwe points. When the absorbance of a solution of unknown concentration (but same substance) is determined, find the concentration from the line at that absorption value. See attached graph.
E.g., an sample of the same substance had an absorbance of 0.35. Find that on the x scale and then determine the concentration that would be required to produce that level of absorbance. 0.483M in this case.
D
is it select all that work or just one?