All categories

1 year ago

Which of the following will affect the vapor pressure of a pure molecular substance? Select all that apply.

1 answer:

4 0

The vapour of a pure chemical substance depends on the level of a intermolecular forces, that substance's structure, and the temperature.

<h3>What are instances of intermolecular forces?</h3>

Between molecules, intermolecular forces are at work. In contrast, molecules themselves exert intramolecular pressures. In comparison to intramolecular forces, intermolecular forces are weaker. The Weaker intermolecular forces, dipole-dipole interaction, ion-dipole communication, and van der Waals forces are a few examples of intermolecular forces.

<h3>What causes intermolecular forces?</h3>

Electrostatic in nature, forces between molecules result from the interaction of positive and negative charges entities. Forces between molecules are the total of both attracting and repulsive elements, similar to held together by covalent connections.

To know more about intermolecular forces visit:

brainly.com/question/9007693

#SPJ4

You might be interested in

The explosions that take place during a fireworks show are an example of which of the following?Image of fireworks exploding in

babymother [125]

Chemical change
:)
hope this helps

4 0

3 years ago

Read 2 more answers

14 protons,14 electrons and 14 neutrons

vovikov84 [41]

The answer is silicon.

6 0

3 years ago

Read 2 more answers

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

For the reaction N2O4(g) ⇋ 2NO2(g), Kc = 0.25 at 98°C. At a point during the reaction, the concentration of N2O4 = 0.50 M and th

Scilla [17]

Answer:

Q = 0.50

Left

Explanation:

At a generic reversible equation

aA + bB ⇄ cC + dD

The reaction coefficient (Q) is the ratio of the substances concentrations:

$Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}$

Solids and liquid water are not considered in this calculus.

When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.

In this case:

Q = $\frac{[NO_2]^2}{[N_2O_4]}$