All categories

2 years ago

At what temperature is a gas likely to have the lowest rate of diffusion? 15° 21° 24° 30°

Chemistry

2 answers:

sergiy2304 [10]2 years ago

8 0

Answer:

At 15° the gas will have lowest rate of diffusion.

Explanation:

Temperature directly affects the kinetic energy of gas molecules. The energy of motion is called kinetic energy. So if we increase the temperature of gas the average kinetic energy of the gas molecules will increase (they will move faster). Due to increase in kinetic energy they will easily diffuse from their higher concentration to lower concentration (which is diffusion).

Hence the rate of diffusion at given temperature will have following increasing order.

15° < 21° < 24° < 30°

Ray Of Light [21]2 years ago

7 0

AT 15°

The rate of diffusion of a gas is directly proportional to the absolute temperature. At lower temperature the gas particles have less kinetic energy than at higher temperature. This makes the particles travel faster at higher temperature than lower temperature.

You might be interested in

Explain how Milankovitch cycles can be used to predict future changes in Earth's climate

Strike441 [17]

Using deep-sea sediment cores found that Milankovitch cycles correspond with periods of major climate change over the past 450,000 years, with Ice Ages occurring when Earth was undergoing different stages of orbital variation.

6 0

2 years ago

From the chair conformation, rotate about the carbon-carbon bonds of the ring to form a boat conformation. In the boat conformat

pychu [463]

Is there a graph? I need to see one

5 0

2 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

2 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

3 0

2 years ago

PLEASE HELP 30 POINTS

posledela

The answears are in the attached photo.

4 0

2 years ago

Read 2 more answers