All categories

2 years ago

Which processes represent one chemical change and one physical change

Chemistry

1 answer:

Effectus [21]2 years ago

6 0

- freezing and melting

- freezing and vaporization

- decomposition and melting

- decomposition and combustion

PLEASE RATE AS THE BRAINLIEST ANSWER! THANK YOU! :)

You might be interested in

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

2 years ago

Which scientist performed the cathode ray experiment leading to the discovery of electrons

SVETLANKA909090 [29]

The scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.

<h3>What is a cathode ray?</h3>

A cathode ray is a tube that contains negatively charged electrode( that is the cathode) which emits electrons when heated at a low pressure.

The cathode ray was used by the scientist, J.J. Thomson to find the ratio of charge to mass (e/m) of the electrons.

Therefore, the scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.

Learn more about electrons here:

brainly.com/question/11316046

#SPJ1

6 0

1 year ago

Stolb23 [73]

Answer:

Um english please I don’t know Jewish

Explanation:

6 0

2 years ago

What is the average kinetic energy of any substance at 0 K?

Aloiza [94]

The answer is zero!

let me know if you need help with anything else! :)

5 0

2 years ago

Read 2 more answers

A student makes the following two claims about chemical reactions:

Svet_ta [14]

I would agree with the second one, not the first. You can't always see the chemical reaction, and it isn't always sudden. But the second claim is true.

7 0

3 years ago