Draw a complete structural formula and a condensed structural formula for(a) three compounds of formula C3H8O (b) five compounds
of formula C3H6O
1 answer:
Answer:
Explanation:
A. Attached is a page which contain the structural formula of the three compounds of C3H80.
Condensed structural formula of C3H8O:
Propan-1-ol: CH3CH2CH2OH
Propan-2-ol: CH3CH(OH)CH3
Methoxy methane: CH3OCH2CH3
B. Attached are is a page which contain the structural formula of the three compounds of C3H60.
Condensed structural formula of C3H6O:
Propanal: CH3CH2CHO
Propanone: CH3COCH3
Cyclopropanol: (C3H5)OH
2-propen-1-ol: CH2CHCH2OH
1-propenol: CH3CHCHOH
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Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄
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:
- NaOH: 3 moles
- H₃PO₄: 1 mole
- H₂O: 3 moles
- Na₃PO₄: 1 mole
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?
moles of NaOH= 27 moles
But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.
<h3>Moles of Na₃PO₄ formed</h3>Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?
<u><em>moles of Na₃PO₄= 2 moles</em></u>
Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.
Learn more about the reaction stoichiometry:
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Answer: 318 K
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 231 kPa
= final pressure of gas = 168 kPa
= initial volume of gas = 3.25 L
= final volume of gas = 4.35 L
= initial temperature of gas =
= final temperature of gas = ?
Now put all the given values in the above equation, we get:
At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa