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3 years ago

13

The following molecular equation represents the reaction that occurs when aqueous solutions of silver nitrate and aluminum chlor

ide are combined. 3 AgNO3(aq) + AlCl3(aq) 3 AgCl(s) + Al(NO3)3(aq) Write the balanced net ionic equation for the reaction.

1 answer:

irga5000 [103]3 years ago

3 0

Answer:

3Ag^(+) + 3Cl ^(-) --> 3AgCl (s)

Explanation:

Full ionic equation:

3Ag^(+) + 3NO3^(-) + Al^(+3) + 3Cl^(-)-->

3AgCl (s) + Al^(3+) + 3NO3^(-)

Spectator ions 3NO3^(-) & Al^(+3) can be left out in the net ionic equation.

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As the distance between the moon and Earth increases the moon's orbital speed...

AleksAgata [21]

Although the moon's distance from earth varies each month because of its eccentric orbit, the moon's mean distance from Earth is nonetheless increasing at the rate of about 3.8 centimeters (1.5 inches) per year

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3 years ago

I. Witch direction does the electric field point at a position directly west of a positive charge

RUDIKE [14]

Answer:

The answer is North

Explanation:

The direction of the field is taken to be the direction of the force it would exert on a positive test charge.

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3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

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:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

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

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

Problem Page Question It takes to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbo

Marizza181 [45]

This is a incomplete question. The complete question is:

It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits

Answer: 344 nm

Explanation:

$E=\frac{Nhc}{\lambda}$

E= energy = 348kJ= 348000 J (1kJ=1000J)

N = avogadro's number = $6.023\times 10^{23}$

h = Planck's constant = $6.626\times 10^{-34}Js
$

c = speed of light = $3\times 10^8ms^{-1}$

$348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}$

$\lambda=3.44\times 10^{-7}m=344nm$ $1nm=10^{-9}m$

Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm

5 0

3 years ago

Which identifies an oxidation-reduction reaction? a double replacement reaction a neutralization reaction a reaction in which ox

OLga [1]

A reaction in which oxidation numbers change is the answer! :D

☆ Dont forget to mark brainliest ☆

7 0

3 years ago

Read 2 more answers