All categories

3 years ago

True or false. Gas particles never touch each other.

Chemistry

1 answer:

Sati [7]3 years ago

6 0

Answer:

False. In a gas, particles are in continual straight-line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other.

Explanation:

Hope this helps! :)

You might be interested in

Tuliskan persamaan tetapan kesetimbangan untuk reaksi-reaksi berikut a. Fe3+(aq) + SCN– (aq) ↔ FeSCN3+(aq) b. 3Fe(s) + 4H2O(g) ↔

nevsk [136]

Answer:

a. $K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}$

b. $K_p = \dfrac{[H_2]^4}{[H_2O]^4}$

Explanation:

Untuk semua jenis reaksi umum:

$aA + bB\iff cC + dD$

Konstanta kesetimbangan $K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}$

Dari pertanyaan yang diberikan:

a. $Fe3^+_{(aq)} + SCN^-_{ (aq)} \iff FeSCN^{3+}_{(aq) }$

Konstanta kesetimbangan:

$K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}$

b. $3Fe_{(s)} + 4H_2O_{(g)} \iff Fe_3O_4_{(s)} + 4H_{2(g)}$

Konstanta kesetimbangan untuk tekanan parsial $K_p$

$K_p = \dfrac{[H_2]^4}{[H_2O]^4}$

Karena Fe3O4 (s) hadir sebagai padatan.

8 0

2 years ago

How does concentration affect the chemical equilibrium?

nevsk [136]

When the concentration of a reactant is increased, the chemical equilibrium will shift towards the products. More product is formed and the concentration of the reactants decreases as the concentration of the products increases.

7 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

3 years ago

"which is a correct description of the organization of subatomic particles in atoms?"

Irina18 [472]

Which is a correct description of the organization of subatomic particles in atoms?

Protons and neutrons are tightly packed into a small nucleus. Electrons occupy the space
outside the nucleus

hope this helps.

5 0

3 years ago

Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

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

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

5 0

3 years ago