All categories

3 years ago

To convert between moles and molecules, you need to...

Chemistry

1 answer:

Kobotan [32]3 years ago

6 0

i bief it is d. halp you at all

You might be interested in

Write out the balanced equation for the following word equation. Use the proper state abbreviations for each of the reactants an

skad [1K]

Answer:

Li (s) + H₂O (l) → 1/2 H₂ (g) + LiOH (aq)

Explanation:

Lithium metal reacts with liquid water to form hydrogen gas and aqueous lithium hydroxide

2Li (s) + 2H₂O (l) → H₂ (g) + 2LiOH (aq)

We can divide stoichiometry coefficients /2, to have the lowest stoichiometry.

3 0

3 years ago

1.) Fluorite has a glassy luster and a white streak. It's an inorganic solid with a chemical formula of CaF2 CaF2

Nesterboy [21]

Answer:

D. All of the above

Explanation:

8 0

3 years ago

Read 2 more answers

What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.

OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The 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 $E^o{cell}$ 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:

$E^o_{cell}=(+0.34V)-(-0.76V)$

$E^o_{cell}=1.1V$

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}$

$E_{cell}=1.14V$

Therefore, the voltage of the voltaic cell is 1.14 V

5 0

3 years ago

HELPPP!!!

dedylja [7]

Hydropower is one of the oldest renewable resources and has been used for thousands of years.

5 0

2 years ago

Can the groups of the periodic table be numbered in two different ways

Vadim26 [7]

The groups (left to right) are numbered from 1 to 18. they arent numbered any other way.

period (up to down.) are numbered from 1-7

3 0

3 years ago