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Lelu [443]
3 years ago
11

How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this

temperature is 31.0 kJ/molkJ/mol.
Chemistry
1 answer:
KiRa [710]3 years ago
4 0

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = \frac{mass}{molar mass}

Number of moles of acetone = \frac{31.5}{58.08}

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

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Deffense [45]

The unhybridized pz orbitals on each carbon overlap to a π bond (pi).The sigma bond framework of the ethylene molecule is produced by the overlap of hybrid orbitals or by the interaction of a hybrid orbital and a 1s hydrogen orbital.

Each carbon still has its unhybridized pz orbital, though. Sigma bond are typically the only types of single bonding between atoms. One sigma bond and two pi bonds make up triple bonds. One sigma () bond makes up a single bond, one and one pi () bond makes up a double bond, and one and two bonds make up a triple bond.

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2 years ago
2. What is the weight of the hydrochloric acid that fills a 144.5 mL container? The density of
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I don’t understand the question sorry
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3 years ago
When examining features on the ocean floor, like a volcano, how would it appear?
kogti [31]

The correct answer is 33% i tried it on a lesson.

8 0
3 years ago
What is the molarity of the Ca(OH)2 solution if 32.00 mL of Ca(OH)2 requires 16.08 mL of a 2.303 M solution for complete titrati
amm1812

Answer:

For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb

  Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L

 

  So now we can plug into the equation and solve:

   1.338 mol/L * 0.01875 L= x mol/L * 0.02473 L

This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.

  First step: setup the stoichiometry problem:

  3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl

 

  Second step: Molarity equation

  Molarity= moles HCl/ L   M= mol HCl/0.04027 L

For the third problem, you will just use the same equation as the first: MaVa= MbVb

  Ma=0.57 M   Va= x L   Mb= 0.875 M  Vb= 23.83 mL= 0.02383 L

  0.57 M * x L= 0.875 M * 0.02383 L

With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.

   0.75 M= x mol/ 0.0227 L    mol NaOH = 0.75 M *0.0227 L         mol NaOH= 0.017025

  So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles

  Molar Mass= 3.6 g/ 0.017025 mol  

 And with that you will find the molar mass of HX, and even determine what X is.

Explanation:

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8 0
2 years ago
Calculate the hydronium ion concentration in an aqueous solution with a poh of 4.33 at 25°c.
Jlenok [28]

Taking into account the definition of pH and pOH, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or  hydronium ion H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of hydronium ions</h3>

Being pOH= 4.33, pH is calculated as:

pH + 4.33= 14

pH= 14 - 4.33

<u><em>pH= 9.67</em></u>

Replacing in the definition of pH the concentration of hydronium ions is obtained:

- log [H₃O⁺]= 9.67

Solving:

[H₃O⁺]= 10⁻⁹ ⁶⁷

<u><em>[H₃O⁺]= 2.138×10⁻¹⁰ M</em></u>

Finally, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

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