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IgorLugansk [536]

3 years ago

9

Answer the table please

1 answer:

fiasKO [112]3 years ago

7 0

B=20 km

C=80km

D=130km

E= 200km

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What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso

Paraphin [41]

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

6 0

3 years ago

What causes the high luster of a metal?

Karo-lina-s [1.5K]

Answer:

When light is shone on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals. The electrons immediately fall back down to lower energy levels and emit light. This process is responsible for the high luster of metals.

Explanation:

<em> </em><em>Your </em><em>well-wisher</em><em> </em><em>:-)</em>

7 0

3 years ago

A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:

seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

$3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)$

<u>Oxidation half reaction:</u> $Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V$ ( × 2)

<u>Reduction half reaction:</u> $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V$ ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=0.16-(-1.66)=1.82V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = $42^oC=[42+273]K=315K$

F = Faraday's constant = 96500

$[Al^{3+}]=1.63M$

$[Cu^{2+}]=3.43M$

Putting values in above equation, we get:

$E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V$

Hence, the cell voltage of the given reaction is 1.86 V

4 0

3 years ago

If the percent (mass/mass) for a solute is 16% and the mass of the solution is 400 g,

Svetllana [295]

Answer:

$m = 64\,g$

Explanation:

The mass of the solute is given by the following expression:

$m = 0.16\cdot (400\,g)$

$m = 64\,g$

4 0

4 years ago

Is this equation balanced? Why or why not?

AnnyKZ [126]

No the equation is not balanced.

5 0

4 years ago

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