All categories

3 years ago

What is the balance of 2C2H6 + O2 → 6H2O + 4CO2

Chemistry

2 answers:

ddd [48]3 years ago

7 0

The balance is 4c02. :D

Hope I helped!

katrin2010 [14]3 years ago

4 0

The balance of the equation is 4c02

You might be interested in

El aguacate al oxidarse, ¿que fenómeno es?, ¿físico, químico o alotropicos?

devlian [24]

Explanation:

WE A është ai Në The H Përshëndetje Nme është

3 0

3 years ago

What are differences and similarities between pure substances and mixtures

Snezhnost [94]

Answer:

HELLO THERE!

I HOPE MY ANSWER WILL HELP YOU :)

Explanation:

There is this picture that helped me. I hope it helps you too.

5 0

3 years ago

Rank the following solutions in order of increasing acidity (least acidic to most acidic): I: [H3O+] = 1x10−5 II: [OH−] = 1x10−1

kozerog [31]

Answer: option C) II < III < I

i.e [OH−] < [H3O+] < I

Explanation:

First, obtain the pH value of I and II, then compare both with III.

For I

Recall that pH = -log (H+)

So pH3O = -log (H3O+)

= - log (1x10−5)

= 4

For II

pOH = - log(OH-)

= - log(1x10−10)

= 9

For III

pH = 6

Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4

Thus, the following solutions from least acidic to most acidic is II < III < I

6 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

3 years ago

An irreversible change would be?

klemol [59]

Answer:

A. Chipping ice to flakes

Explanation:

You can't reverse the ice becoming flakes. They will stay like that until they melt

8 0

3 years ago