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3 years ago

URGENT!!!

Chemistry

1 answer:

Inga [223]3 years ago

6 0

Answer:

${ \bf{from \: ionic \: product \: of \: water : }} \\ { \boxed{ \tt{k _{w} = [H _{3} O {}^{ + } ][OH {}^{ - } ]}}} \\ \\ { \tt{1 \times {10}^{ - 14} = (1 \times {10}^{ - 5} ) [OH {}^{ - } ]}} \\ \\ { \tt{[OH {}^{ - } ] = 1 \times {10}^{ - 9} }} \: M$

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Energy is transferred

Andrei [34K]

True is the right oneee

6 0

3 years ago

Which is a property of barium (Ba)?

777dan777 [17]

Answer:

i think but i am not sure but according to me it mainly reacts yo non metals and i think its very reactive if my answer is wrong then comment below this question i will see it and i will get an opportunity to learn something new

4 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

A particular brand of gasoline has a density of 0.737 g/ml at 25 ∘c. how many grams of this gasoline would fill a 13.0 gal tank

Usimov [2.4K]

Density of gasoline is 0.737 g/mL and volume of tank is 13.0 gal.

Since, 1 US gal=3.78 L

Volume of tank in L will be:

$V=13.0 gal(\frac{3.78 L}{1 gal})=49.14 L$

Also, 1 L=1000 mL

Thus,

$V=49.14 L(\frac{1000 mL}{1 L})=49140 mL$

Mass of gasoline can be calculated as follows:

m=d×V

Here, d is density and V is volume thus,

$m=0.737 g/mL\times 49140 mL=3.62\times 10^{4}g$

Therefore, mass of gasoline will be $3.62\times 10^{4}g$ .

7 0

3 years ago

Which of the following measurements (of different masses) is the most precise? A 6.2000g B 6.000000g C 6.24400g D 6.24444g

babymother [125]

Answer:

Explanation:

7 0

3 years ago