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3 years ago

8

How did planetesimals form?

2 answers:

diamong [38]3 years ago

5 0

The process is still not fully understood.
<span>Planetesimals formed from dust and gas in the "protoplanetary disk". </span>
<span>Electrical attraction seems to be important in the early stages then gravity </span>
<span>brings things together. </span>
<span>The electrical forces are van der Waals forces.</span>

Nady [450]3 years ago

3 0

Condensation, the process of forming solid particles from the solar nebula, is an early process in the formation of a planetesimal, an early solar system body of small to medium size that combines with otherplanetesimals<span> to </span>form<span> protoplanets</span>

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NO<br> Assign oxidation numbers to each element in this compound.

marta [7]

Answer:

N=+2

O=-2

Explanation:

The compound NO is electrically neutral.

Lets assign the oxidation number of nitrogen to be N. The oxidation number of oxygen (-2) is then used as a reference.

For the compound to have a zero charge, sum of the oxidation numbers equals zero.

N+ (-2)=0

N=+2

O=-2

7 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as 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:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

In reversible reaction when do forward reactions take place

kari74 [83]

Answer:

that is true. because reversible react is take place in forward and backward reaction

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dezoksy [38]

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butalik [34]

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