It’s true:)
The subscript on chemical equations tells you how many atoms the elements have. If there is no subscript then there is only 1 atom.
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Answer:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
Explanation:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:
<u>Where:</u>
<em>μ (l): is the chemical potential of 2-propanol in solution </em>
<em>μ° (l): is the chemical potential of pure 2-propanol </em>
<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>
<em>T: is the temperature = 82.3 °C = 355.3 K </em>
<em>x: is the mole fraction of 2-propanol = 0.41 </em>

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
I hope it helps you!
<span>Based on your information 1000 times greater than pH 13 is the best I can come up </span>with.
Answer : The amount of formaldehyde permissible are, 
Explanation : Given,
Density of air =

First we have to calculate the mass of air.



Now we have to calculate the amount of formaldehyde.
Permissible exposure level of formaldehyde = 0.75 ppm = 
Amount of formaldehyde in 7.2 g of formaldehyde = 
Amount of formaldehyde in 7.2 g of formaldehyde = 
Thus, the amount of formaldehyde permissible are, 