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4 years ago

Help me please ...online claases got me crazy

Chemistry

1 answer:

White raven [17]4 years ago

6 0

Answer:

1,000 - 2,000

Explanation:

Just look at the bar graph. The bar ends at about 1,500 which is between 1,000 - 2,000. That is the only valid answer for this problem.

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Periodic table the first row of elements fits in the period blank after the element blank the second row of elements fits in per

Bezzdna [24]

Answer:

The first row fits in period 2

The second row fits in period 3

Explanation:

Elements arranged vertically in the periodic table in groups that share similar chemical properties.

Elements are also organized horizontally in rows or periods

The length of each period is determined by the number of electrons that can occupy the sublevels being filled in that period.

7 0

3 years ago

HEEEEEEEEEEEEEEELLLLLLLLLLLLPPPPPPP!!!!!!! I WILL MARK BRAINLIEST IF YOU HELP!!!!THIS IS DUE SOON!

Hitman42 [59]

Answer:

An ice pack uses friction and the chemicals inside make it mix to make it cold

Explanation:

3 0

3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,

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

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,

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

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

How does Gibbs free energy predict spontaneity?

Kaylis [27]

Answer:

Gibbs equation helps us to predict the spontaneity of reaction on the basis of enthalpy and entropy values directly. When the reaction is exothermic, enthalpy of the system is negative making Gibbs free energy negative. Hence, we can say that all exothermic reactions are spontaneous.

5 0

3 years ago

Tell me why in Louisiana it always gotta freaking have a hurricane

Gala2k [10]

Why do so many storms target the state? The easterly winds typically blowing across Jacksonville in the summer flow out of the Azores/Bermuda high-pressure system which also steers many hurricanes into the Gulf. In June and October, storms are more likely to move up the Gulf from the south and southwest.

7 0

3 years ago