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3 years ago

Two answer multiple-choice-- please help!

Chemistry

1 answer:

Troyanec [42]3 years ago

6 0

Answer:

A spray of perfume diffusing in a room is a spontaneous process because gas particles have higher entropy when they are spread evenly around the room.

This is true because entropy increase when gas particles spread

Explanation:

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3. Round 87.073 meters to three significant figures. Write your answer in scientific notation. Step 1: 87.073 rounds to _______

Nat2105 [25]

Answer:

see below

Explanation:

3. 87.073 meters to three significant figures.

= 87.1 meters

2. Write the number in scientific notation. _____meters

= 8.71 x 10 ¹ meters

6 0

3 years ago

Read 2 more answers

Any two uses of mixtures

romanna [79]

Answer:

Any two used of mixtures are 1) sand and water 2) sugar and salt

Explanation:

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3 years ago

On Earth, the mass of one of the lunar landing modules used to explore the moon was measured to be about 4,200 kg. The accelerat

Fed [463]

7 0

4 years ago

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1 C3H8 + 5 O2 --> 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many

UNO [17]

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0

4 years ago

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago