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3 years ago

Which equation represents the combined gas law?

Chemistry

2 answers:

aksik [14]3 years ago

7 0

The correct option is P1V1 / T1 = P2V2 /T2.

The combine gas law states that the ratio of the product of pressure, volume and absolute temperature of a gas is always a constant. Mathematically, the law is represented by P1V1 / T1 = P2V2 /T2; where

P1 and P2 = Initial and final pressure of the gas

V1 and V2 = Initial volume and the final volume of the gas

T1 and T2 = Initial temperature and final temperature of the gas.

Combine gas law was formed by the combination of Boyle's law, Charles' law and Gay Lussac's law.

Artist 52 [7]3 years ago

5 0

Explanation :

<em>Combined gas law comes from the combination of Boyle's law, Charles law and Gay Lussac's law.</em>

According to Boyle's law,

At constant temperature, pressure of a gas is inversely proportional to its volume.

i.e. $P\propto\dfrac{1}{V}$

$PV=constant$ ..........(1)

Where,

P is the pressure of gas

V is the volume of gas

According to Charles law,

At constant pressure, volume of a gas is directly proportional to its temperature.

i.e. $V\propto T$

$\dfrac{V}{T}=constant$ ..............(2)

Where, T is the temperature.

According to Gay Lussac's law,

At constant volume, pressure of gas is directly proportional to its temperature.

i.e. $P\propto T$

$\dfrac{P}{T}=constant$ ..............(3)

Combining equation (1) (2) and (3)

$\dfrac{PV}{T}=constant$ ...............(4)

Equation (4) is the combined gas law.

Or, $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

So, correct option is (c).

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kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

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

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

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Identify the neutral element represented by this excited-state electron configuration, then write the ground-state electron conf

sineoko [7]

Answer:

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Explanation:

Excited state configuration: 1s² 2s² 2p² 3s¹

Unknown:

Ground state configuration = ?

Element symbol = ?

Solution:

Let us start by understanding what a ground state configuration entails:

A ground state configuration shows the lowest allowed energy levels of an atom. The excited state denotes when electrons have moved to higher energy levels away from their ground state.

The superscript in the configuration depicts the number of electrons in each of the sublevels.

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In writing the electronic configuration of an atom, certain rules must be complied with.

Aufbau's principle states that the sublevels with lower energies are filled up before those with higher energies. The order of filling is:

1s 2s 2p 3s 3p 4s etc

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The outermost shell electrons are usually the ones excited.

Now, the 3s¹ shell is an excited one taking 1 electron from the second energy level;

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