All categories

3 years ago

Based on your observations, which equation best models the reaction of vinegar (C2H4O2) and baking soda (NaHCO3)? Explain.

Chemistry

1 answer:

tensa zangetsu [6.8K]3 years ago

8 0

Answer: im in 8th they got me doin this

Explanation: i need help quick

You might be interested in

Earth's gravitational potential energy: GPE = mgh = Gravity (9.81m/s2) Mass (kg) x Height (m) Kinetic energy: KE = mu? How do th

Sati [7]

Answer:D

Explanation: i just took the quiz (apex)

5 0

3 years ago

Molecules are the smallest units of matter with the characteristic properties of a substance.

Vilka [71]

False, Atoms are the smallest units of matter that display both the chemical and physical properties of it, based on the structure of the atom.

3 0

4 years ago

What does the term division of labor mean as it relates to cells?

lidiya [134]

Answer:

what each cells job is divided

Explanation:

7 0

3 years ago

How many particles are present in 8.0 moles of silver?

Zielflug [23.3K]

Answer:the answer is b

Explanation:I took the test and got it right

3 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