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Harlamova29_29 [7]
2 years ago
15

Consider the following molecules and the description of the bonding present in each: CH3CH2CH2CH3CH3CH2CH2CH3 (C−CC−C and C−HC−H

bonds only) CH3CH2CH2OH CH3CH2CH2OH (C−CC−C and C−HC−H bonds, and one C−OHC−OH bond) CH3CH2CHO CH3CH2CHO (C−CC−C and C−HC−H bonds, and one C=OC=O bond) Based on the general trends of intermolecular forces, rank these molecules in order of increasing boiling point.
Chemistry
1 answer:
denis23 [38]2 years ago
3 0

CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3

Explanation:

Boiling point trend of Butane, Propan-1-ol and Propanal.

Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.

In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.

Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.

In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol

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Consider the reaction between two solutions, X and Y, to produce substance Z: aX + bY → cZ When 500. mL of a 1.8 M solution of X
nignag [31]

Answer:

1. 0.90 are the initial moles of X and Y

2. 0.60 moles are the moles of Y and Z after the reaction

3. 0.90 moles of X and 0.30 moles of Y

4. 3X + 1Y → 2Z

Explanation:

1. For the reaction, initial moles of X and Y are:

500mL = 0.500L × (1.8 moles / L) = 0.90 are the initial moles of X and Y

2. After the reaction. The total volume is 500mL + 500mL = 1L

Moles Y and Z = 1L × (0.60 moles / 1L) = 0.60 moles are the moles of Y and Z after the reaction

3. As there is no moles of X after the reaction, all X reacts, that is 0.90 moles of X. And moles of Y that reacts are 0.90 mol - 0.60mol = 0.30 moles of Y

4. That means 3 moles of X reacts per mole of Y 0.90/0.30 = 3. Also, 2 moles of Z are produced per mole of Y 0.60/0.30 = 2.

That means balanced equation is:

aX + bY → cZ

<h3>3X + 1Y → 2Z</h3>
7 0
2 years ago
The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard for calibrating constant-volume bomb calo
Anit [1.1K]
<h3>Answer:</h3>

28.96 kJ/°C

<h3>Explanation:</h3>

We are given;

  • Enthalpy change (ΔH) = −3226.7 kJ/mol
  • The reaction is exothermic since the heat change is negative;
  • Mass of benzoic acid = 3.1007 g
  • Temperature change (21.84°C to 24.67°C) = 2.83°C

We are required to find the heat capacity of benzoic acid;

<h3>Step 1: Moles of benzoic acid </h3>

Moles = Mass ÷ molar mass

Molar mass of benzoic = 122.12 g/mol

Therefore;

Moles = 3.1007 g ÷ 122.12 g/mol

          = 0.0254 moles

<h3>Step 2: Determine the specific heat capacity </h3>

Heat change for 1 mole = 3226.7 kJ

Moles of Benzoic acid = 0.0254 moles

But;

Specific heat capacity × ΔT = Moles × Heat change

  cΔT = nΔH

Therefore;

Specific heat capacity,c = nΔH ÷ ΔT

                                     = (3226.7 kJ × 0.0254 moles) ÷ 2.83°C

                                    = 28.96 kJ/°C

Therefore, the specific heat capacity of benzoic acid is 28.96 kJ/°C

7 0
2 years ago
All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou
____ [38]

Answer:

m_{Mg}=30.8mgMg

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg

Best regards.

3 0
1 year ago
The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 375 kJ/mol and the change in enthalpy for the reaction
HACTEHA [7]
Answer: 625 kj/mol

Explanation:

As shown below this expression gives the activation energy of the reverse reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

1) The activation energy, EA is the difference between the potential energies of the reactants and the transition state:

EA = energy of the transition state - energy of the reactants.

2) The activation energy of the forward reaction given is:

EA = energy of the transition state - energy of  [ NO2(g) + CO(g) ] = 75 kj/mol

3) The negative enthalpy change - 250 kj / mol for the forward reaction means that the products are below in the potential energy diagram, and that the potential energy of the products, [NO(g) + CO2(g) ] is equal to 375 kj / mol - 250 kj / mol = 125 kj/mol

4) For the reverse reaction the reactants are [NO(g) + CO2(g)], and the transition state is the same than that for the forward reaction.

5) The difference of energy between the transition state and the potential energy of [NO(g) + CO2(g) ] will be the absolute value of the change of enthalpy plus the activation energy for the forward reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

EA reverse reaction = 375 kj / mol + |-250 kj/mol | = 375 kj/mol + 250 kj/mol = 625 kj/mol.

And that is the answer, 625 kj/mol
4 0
1 year ago
Why are familiar objects such as pens and paper clips not commonly counted in moles?
Anton [14]
Well ask yourself why don't we count it in moles and you should get your answer.
6 0
2 years ago
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