There are only <em>three</em> constitutional isomers (see the image below).
Ethyl methyl ether is the least acidic because it has only C-H bonds. However, it should be more acidic than an alkane (pKₐ ≈ 50) because of the electron withdrawing effect of the O atom. I would expect it to have pKₐ ≈ 30.
The alcohols are <em>more acidic</em> because the O in the O- H bond is highly electron withdrawing. Alcohols typically have pKₐ ≈ 17.
Alkyl groups are electron-donating. This increases electron density in the OH bond and makes the acid weaker. Propan-2-ol has two methyl groups on the α-carbon, while propan-1-ol has one ethyl group. Thus propan-2-ol should be a <em>slightly weaker</em> acid than propan-1-ol.
Answer:
0.4383 g
Explanation:
Molality is defined as the moles of the solute present in 1 kg of the solvent.
It is represented by 'm'.
Thus,
Given that:
Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )
Molality = 0.050 m
So,
Molar mass of NaCl = 58.44 g/mol
Mass = Moles*Molar mass = = 0.4383 g
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?
<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
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