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Masja [62]
3 years ago
14

Outline the difference between energy and matter

Chemistry
1 answer:
astra-53 [7]3 years ago
3 0
Matter is everything and anything that has mass and takes up space. Energy<span> is the ability to cause change or do work... Energy, basically, has the power to make a difference.  
</span>
 Energy<span> examples:
 light, heat, </span>nuclear, chemical, and electrical/mechanical energy.

Matters examples:
A pencil, milk, and steam.

I hope this helps!

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What is the apparatus used to carry out paper chromatography?
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Answer:

The apparatus used for paper chromatography are strip of spotted whatman filter paper, Organic solvents, paper strip, trough for solvent, Hook to hold the chromatography paper.

Explanation:

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Approximately how much of the world’s oil and natural gas reserves are believed to be in the arctic? View Available Hint(s) Appr
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1/4 or 25%

Explanation:

The Arctic region of the earth refers to that part of the earth around the north pole region. Hence, when we are talking about latitude O degrees North, the areas around this geographical location is referred to as the arctic.

Now , there is an estimated 1/4 or 25% of the world’s oil and natural gas here. Unfortunately, these are yet accessible because of the amount of ice or snow covering. With increase in technological advancements, this might be accessible in the nearest future

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3 years ago
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Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10
mylen [45]

<u>Answer:</u> The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

<u>Explanation:</u>

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

                   C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)

<u>Initial:</u>          24.0

<u>At eqllm:</u>    24-x            x              x

The expression of K_p for above equation follows:

K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}

We are given:

K_p=0.040

Putting values in above expression, we get:

0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

PV=nRT          .........(1)

To calculate the mass of a substance, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}           ..........(2)

  • <u>For ethane gas:</u>

We are given:

P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}

Putting values in equation 1, we get:

23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g

  • <u>For ethylene gas:</u>

We are given:

P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}

Putting values in equation 1, we get:

0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g

  • <u>For hydrogen gas:</u>

We are given:

P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}

Putting values in equation 1, we get:

0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

5 0
3 years ago
A toaster has a resistance of 28 Ω and is plugged into a 110 V circuit. Using the GRASS method, what current is measured?
BARSIC [14]

Answer:

The current of the toaster is 3.93A

Explanation:

Using the GRASS method:

G: Given: The information that is given is that we have 28Ω of Resistance and 110V of voltage.

R: Required: We need to find the current measured.

A: Action: We can use Ohm's law that relates resistance, voltage and current in the equation:

V = R*I

Or:

I = V/R

<em>Where I is current in A, V is voltage = 110V and R is resistance = 28Ω</em>

S: Solve:

I = 110V / 28Ω

I = 3.93A

S: Statement:

<h3>The current of the toaster is 3.93A</h3>
4 0
3 years ago
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