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3 years ago

What is the relationship between the perimeter and the side length of anequilateral triangle ?

Physics

1 answer:

katovenus [111]3 years ago

3 0

Answer:

The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90. Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

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3 years ago

Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction

lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,

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

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

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

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

$E^o_{[Ag^{+}/Ag]}=0.80V$

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

$E^o=0.80V-(0.34V)=0.46V$

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

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

where,

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

$E_{cell}$ = ?

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

$E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}$

$E_{cell}=0.434V$

Therefore, the cell potential for this cell 0.434 V

8 0

4 years ago

The area of the bar over r = 2 is 0.234. what is the area of the bar over r = 4?

natita [175]

The area of the bar over r=4 is 0.468

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3 years ago

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Answer:

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mamaluj [8]

Answer:

its b

Explanation:

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