Ask question

Ask question

All categories

Alex_Xolod [135]

3 years ago

10

You have 1 1/2 moles of 1 kg bottles of O2. What is the mass of O2 that you have? A. 9.033x10^23 kg B. 9.033x10^23 atoms C. 1.80

x10^27 kg D. 1.806x10^24 moles

2 answers:

ki77a [65]3 years ago

8 0

Answer:

B

Explanation:

Rashid [163]3 years ago

6 0

Answer:

b

Explanation:

trust

You might be interested in

If you have 1kg of carbon and 1 kg of gold, they will weigh the same.

Mekhanik [1.2K]

This would be be true

8 0

2 years ago

How many moles are in 19.82 g Mg?

vekshin1

I think it’s 24.305? Not sure

4 0

3 years ago

DNA replication results in two identical DNA molecules. What role do DNA helicases have in DNA replication?

PilotLPTM [1.2K]

I'm pretty sure since one chemical can only bind with one other type, um, I think that one side of the DNA helicase helps make the other side?

8 0

3 years ago

Read 2 more answers

An endothermic reaction _____.

Over [174]

Release less energy than it uses

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