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2 years ago

What intermolecular forces are present in each of the substances?

2 answers:

4 0

What intermolecular forces are present in each of the substances? CH4,C3H8,CH3F,HF, C6H5OH (dispersion forces<span>, </span>dipole-dipole forces<span>, or </span>hydrogen bonding<span>);A sample of ideal gas at room temperature occupies a volume of 34.0L at a pressure of 782torr .</span>

pochemuha2 years ago

3 0

What intermolecular forces are present in each of the substances? (1) Chloroform, CHCl3, (2)Oxygen, O2, (3)Polyethylene, CnH2n+2, and (4)Methanol, CH3OH.

<h3>Explanation: </h3>

Intermolecular forces are the forces that mediate interaction between molecules including forces of attraction or repulsion act between molecules and other types of neighboring particles, atoms or ions. The four intermolecular forces are: Ionic bonds, Hydrogen bonding, Van der Waals dipole-dipole interactions, Van der Waals dispersion force. The characteristics are Ionic bonds > Hydrogen bonding > Van der Waals dipole-dipole interactions > Van der Waals dispersion force

An ionic bond is chemical bond type where one or more electrons are transferred from one atom to another. Hydrogen bond is attractive (dipole-dipole) interaction between an electronegative atom and a hydrogen atom bonded to another electronegative atom. Whereas Van der Waals is distance-dependent interaction between atoms or molecules. There are two kinds of Van der Waals forces: weak London Dispersion Forces and stronger dipole-dipole forces.

(1) Chloroform, CHCl3, are intermolecular forces that present in each of the substances
(1) Chloroform, CHCl3: Dipole-dipole intermolecular forces are most important.
(2)Oxygen, O2: London dispersion forces are most important.
(3)Polyethylene, CnH2n+2: London dispersion forces are most important.
(4)Methanol, CH3OH: Dipole-dipole intermolecular forces and hydrogen bounds are important

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What role does the activation energy play in chemical reactions?

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The activation energy is the minimum amount of energy that particles must have in order for them to participate in a chemical reaction. During chemical reactions bonds are broken and formed. Particles must collide with sufficient energy in order for the initial bonds to be broken. The activation energy is that that initial minimum energy that the particles can have in order for the bonds to be broken. Stronger bonds would require more energy to be broken and therefore the activation energy for such would be higher.

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3 years ago

(06.03 MC) A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What i

oee [108]

Answer:

New volume is 25.0 mL

Explanation:

Let's assume the gas sample behaves ideally.

According to combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where, $P_{1}$ and $P_{2}$ represent initial and final pressure respectively

$V_{1}$ and $V_{2}$ represent initial and final volume respectively

$T_{1}$ and $T_{2}$ represent initial and final temperature (in kelvin) respectively

Here, $T_{1}=T_{2}$ , $V_{1}=50.0mL$ , $P_{1}=20.0atm$ and $P_{2}=40.0atm$

So, $V_{2}=\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}=\frac{(20.0atm)\times (50.0mL)}{40.0mL}=25.0mL$

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2 years ago

A particular sample of pure iron requires 0.612 kJ of energy to raise its temperature from 30.°C to 51°C. What must be the mass

mixas84 [53]

Answer:

m = 65.637 g

Explanation:

Heat = 0.612 kJ = 612 J ( Converting to J by multiplying by 1000)

Initial Temperature = 30.°C

Final Temperature = 51°C

Temperature change = Final Temperature - Initial Temperature = 51 - 30 = 21°C

Mass = ?

The relationship between these quantities is given by the equation;

H = mCΔT

where c = 0.444 J/g°C

Inserting the values in the equation;

612 = m * 0.444 * 21

m = 612 / (0.444 * 21)

m = 65.637 g

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2 years ago

Write .. hot copper metal reacts with chlorine gas to form solid green copper chloride

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Answer:

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2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

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2 years ago