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3 years ago

What’s a black dwarf

Chemistry

2 answers:

Helga [31]3 years ago

7 0

Answer:

you

Explanation:

Elena-2011 [213]3 years ago

5 0

Answer: htp its a vid game

Explanation:

Since the theoretical black dwarf is just a white dwarf that has cooled completely, then it should be the same composition as a white dwarf. The final end product of fusion is iron, therefore a black dwarf would be made of iron

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Dessert biomes have all of these EXCEPT-

Dmitry_Shevchenko [17]

Answer: B). little wind

Explanation: Deserts have few large animals, fertile soil, and little rainfall but they have lots of wind

7 0

3 years ago

Can someone please help me fill this out?

Mamont248 [21]

The states of matter is solid,liquid and gas & you go from there.

6 0

3 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

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

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) Now we have to calculate the potential of a cell.

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

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

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

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

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

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

where,

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

$E_{cell}$ = emf of the cell = ?

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

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

5 0

2 years ago

A precipitation reaction is caused by mixing 100. mL of 0.25 M K2Cr2O7 solution with 100. mL of 0.25 M Pb(NO3)2 solution. When t

Sergeeva-Olga [200]

Answer:

Neither is affected

Explanation:

The reaction occurs as follows:

K₂Cr₂O₇ + Pb(NO₃)₂ → PbCr₂O₇ + 2K⁺ + 2NO₃⁻

That means per mole of reaction you will have two moles of both K⁺ and NO₃⁻.

But volume is also doubled, doing that concentration of spectator ions doesn't change.

Right answer: Neither is affected

I hope it helps!

8 0

3 years ago

What is the best explanation of why the iron found in the inner core of the Earth remains in the solid state even though the tem

Musya8 [376]

Answer: b

Explanation: the pressure from gravity in the inner core prevents the iron from melting to a liquid

8 0

3 years ago