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geniusboy [140]
3 years ago
8

What kind(s) of intermolecular forces exist in the compounds given below? (a) CH3CH2CH2CH2CH3(l) dispersion forces dipole-dipole

forces hydrogen bonding (c) H2CO(l) dispersion forces dipole-dipole forces hydrogen bonding (b) CH3CH2OH(l) dispersion forces dipole-dipole forces hydrogen bonding (d) O2(l) dispersion forces dipole-dipole forces hydrogen bonding
Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
7 0

Answer:

(a) CH3CH2CH2CH2CH3(l)

  • dispersion

(b) CH3CH2OH(l)

  • Dipole-dipole interaction
  • Dispersion forces
  • Hydrogen bonding

(c) H2CO(l)

  • Dipole-dipole
  • Dispersion

(d) O2(l)

  • Dispersion

Explanation:

Dispersion forces are those forces that occur between two non polar molecules.They form the weakest bonds.Here electrons of one molecule is attracted to the nucleus of the other molecule. Example are;

  • Interaction of two methyl (-CH₃) group
  • Interaction between nitrogen gas , N₂ molecules
  • Interaction between oxygen gas ,O₂ molecules

Dipole-Dipole interaction happens when two polar molecules interact.Positive charges in one molecule is attracted to negative charge of another molecule.Examples

  • Chloroform (CHCl₃)
  • Ammonia (NH₃)

Hydrogen bonding are created when an hydrogen atom bonded to an electronegative atom is attracted to a another electronegative atom.Example is the hydrogen bond between oxygen atom and hydrogen is water (H₂O).

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A chemist has some 40% acid solution, some 60% acid solution, and a wholebunch of free time. How many liters of each should be u
TiliK225 [7]

Answer:

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

Explanation:

Let volume of the 40% acid solution be x.

Let volume of the 60% acid solution be y.

Volume of solution formed after mixing both solution = 40 L

x + y = 40 L..[1]

Volume of acid 40% solution = 40% of x= 0.4x

Volume of acid 60% solution = 60% of y= 0.6y

Volume of acid formed = 45% of 40 L = \frac{45}{100}\times 40=18L

0.4x+0.6y=18 L..[2]

Solving [1] and [2]

x = 30 L  ,   y = 10 L

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

8 0
3 years ago
What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

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Read 2 more answers
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
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