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I am Lyosha [343]
3 years ago
13

Which is generally stronger, intermolecular interactions or intramolecular interactions? Which is generally stronger, intermolec

ular interactions or intramolecular interactions? Intermolecular interactions are generally stronger. Intramolecular interactions are generally stronger. These interactions are equally strong.
Part B Which of these kinds of interactions are broken when a liquid is converted to a gas? Which of these kinds of interactions are broken when a liquid is converted to a gas?

a. Only intermolecular interactions are broken when a liquid is converted to a gas.
b. Only intramolecular interactions are broken when a liquid is converted to a gas.
c. Both of these interactions are broken.
Chemistry
1 answer:
Inessa [10]3 years ago
7 0

Answer:

A. Intramolecular interactions are generally stronger.

B. a. Only intermolecular interactions are broken when a liquid is converted to a gas.

Explanation:

<em>A. Which is generally stronger, intermolecular interactions or intramolecular interactions?</em>

Intramolecular interactions, in which electrons are gained, lost or shared, constitute true bonds and are one or two orders of magnitude stronger than intermolecular interactions.

<em>B. Which of these kinds of interactions are broken when a liquid is converted to a gas?</em>

When a liquid vaporizes, the intermolecular attractions are broken, that is, molecules get more separated. However, true bonds are not broken which is why the molecules keep their chemical identity.

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The gas left in a used aerosol can is at a pressure of 103 kPa at 25.0 °C. If the can heats up to 50.0 °C, what is the pressure
Anton [14]
The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:     \frac{P_{1} }{T_{1} }    =   \frac{P_{2} }{T_{2} }
Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).
                      \frac{P_{1} }{T_{1} }    =   \frac{P_{2} }{T_{2} }∴ by substituting the known values,                 ⇒       (103 kPa) ÷ (25 °C)  =  (P₂) ÷ (50 °C)
                 ⇒                                  P₂  =  (4.12 kPa · °C) (50 °C)
                                                            =  206 kPa 
Thus the pressure of the gas since the temperature was raised from 25 °C to 50 °C is 206 kPa

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3 years ago
Why does the carbon anode burn away in the electrolysis of aluminium chloride?
Xelga [282]

Answer:

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5 0
2 years ago
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When 3.0 grams of H2 is reacted with excess C at constant pressure, the reaction forms CH4 and releases 53.3 kJ of heat. C(s) +
Murrr4er [49]

Answer:

THE ENTHALPY OF REACTION IN KJ/MOL OF CH4 IS 7.07 KJ/MOL.

Explanation:

Mass of H2 = 3 g

Molar mass of H2 = 2 g/mol

Heat released = 53.3 kJ

Equation of the reaction:

C(s) + 2H2(g) -------> CH4(g)

First:

Calculate the number of moles of H2 that was used:

Number of moles = mass / molar mass

Number of moles = 3g / 2g

Number of moles = 1.5 moles

So therefore, when 53.3 kJ of heat was released from the reaction, 1.5 moles of hydrogen was used.

From the equation of the reaction, one mole of carbon reacts with two moles of hydrogen to form one mole of methane.

For 3 g of hydrogen, 1.5 mole of hydrogen is involved.

It means:

1.5 moles of hydrogen reacts with 0.75 moles of carbon and produces 0.75 moles of methane. This is so because the reaction occurs in 1: 2: 1 in respect to carbon, hydrogen and methane respectively.

So we can say that the production of 0.75 mole of methane will evolve 53.3 kJ of heat.

0.75 mole of methane releases 53.3 kJ of heat.

1 mole of methane will release ( 53.3 kJ * 1 / 0.75 )

= 71.0666 kJ of heat

In conclusion, the enthalpy of the reaction in kJ/ mole of CH4 is 71.07 kJ/mol.

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3 years ago
I WILL GIVE YOU BRAILEST!!!!!!!!!
Sholpan [36]

Answer:

D.

atomic modifier

Explanation:

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What is the pH when the hydrogen ion concentration is 1 x 10-3 M?
Fynjy0 [20]

Answer:

e. 3

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In order to solve this problem we need to keep in mind the definition of pH:

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As stated by the problem, the hydrogen ion concentration, [H⁺], is 1x10⁻³ M.

As all required information is available, we now can <u>calculate the pH</u>:

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The correct option is thus e.

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2 years ago
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