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Arte-miy333 [17]

2 years ago

13

When a sample of oxygen gas in a closed container of constant volume is heated until its absolute temperature is doubled, which

of the following is also doubled?
a. density
b. pressure
c. average velocity
d. number of molecules
e. potential energy

1 answer:

saul85 [17]2 years ago

7 0

Answer: b. pressure

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$ (At constant volume and number of moles)

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure of gas = p

$P_2$ = final pressure of gas = ?

$T_1$ = initial temperature of gas = t

$T_2$ = final temperature of gas = 2t

$\frac{p}{t}=\frac{P_2}{2t}$

$P_2=2p$

Thus the pressure also doubles when absolute temperature is doubled.

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A compound with molecular formula C4H6O4 produces a broad signal between 2500 and 3600 cm-1 in its IR spectrum and produces two

tiny-mole [99]

Answer:

Succinic Acid

Explanation:

We have to start using the info given by the IR spectrum. In this case, we have a <u>broad signal in 3600</u>. This indicates the presence of OH in the structure. Therefore we can have an <u>alcohol or an acid</u> in the structure.

Now, the NMR info tells us that we only have 2 signals, this indicates that we have a <u>very symmetric structure</u>. Also, we have a signal in 12 ppm therefore we can affirm that we have an O-H bond with <u>high polarity</u> (a downfield signal) this behavior is given in the <u>acid groups</u>.

The structure that fulfill these requirements it the succinic acid. In which we only have 2 signals in the 1H- NMR. We have an acid group and we have a formula of $C_4H_6O_4$ .

<u>For further explanations see the attached images.</u>

3 0

2 years ago

15. An apparatus consists of a 4.0 dm3

Solnce55 [7]

Answer:

i· Partial pressure of nitrogen gas is 219.429kPa and partial pressure of argon gas is 33.714kPa ·

ii· Total pressure of the gas mixture is 253.143kPa·

Explanation:

<h3>solution for i :</h3>

Assuming that the given gases to be ideal,

so by ideal gas equation

PV=nRT

where,

P is the pressure of the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Actually partial pressure of each gas is the pressure of the gas exerted when it occupies complete volume

As the temperature of both gases are same, the mixing process is an isothermal process

PV=constant

Initially for nitrogen gas PV=803×4=3212

let P $x_{1}$ be the partial pressure of the nitrogen gas

(P $x_{1}$ )×14=3212

∴P $x_{1}$ =219.429kPa

∴Partial pressure of nitrogen gas is 219.429kPa

Initially for argon gas PV=47.2×10=472

let P $x_{2}$ be the partial pressure of the argon gas

(P $x_{2}$ )×14=472

∴P $x_{2}$ =33.714kPa

∴Partial pressure of argon gas is 33.714kPa

solution for ii :

Total pressure of the gas mixture will be the sum of the partial pressures of each gas as

Partial pressure of the gas=(total pressure of the mixture)×(mole fraction of the gas)

∴Total pressure=P $x_{1}$ +P $x_{2}$

=219.429+33.719

Total pressure =253.143kPa

6 0

2 years ago

Read 2 more answers

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

<u>Answer:</u>

<u>For A:</u> The $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u> The $K_c$ for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

<u>For A:</u>

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

1 year ago

A solid is 5 cm tall 3 cm wide and 2 cm thick it has a mass of 129 g what is the density

belka [17]

First you would multiply 5,3, and 2 to get a volume of 30. Density is mass over volume so you would then divide 129 by 30. You would get 4.3. The answer would be 4.3 g/cm^3

7 0

2 years ago

Read 2 more answers

Why it is impossible for an isolated atom to exist in the hybridized state?

kakasveta [241]

Hybridization refers to the mixing of atomic orbitals in an atom. The number of hybrid orbitals needs to be equal to the number of orbitals that have involved in prior to mixing.

The isolated atoms cannot prevail in a hybridized state as the atom in an isolated state do not form any kind of bond with the other atom, due to which the atomic orbitals do not go through the process of hybridization.

7 0

2 years ago