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soldi70 [24.7K]

3 years ago

8

How do water molecules dissolve a salt?

1 answer:

Artist 52 [7]3 years ago

5 0

Answer:

Water can dissolve salt because the positive part of water molecules attracts the negative chloride ions and the negative part of water molecules attracts the positive sodium ions. The amount of a substance that can dissolve in a liquid (at a particular temperature) is called the solubility of the substance.

Explanation:

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Identify and calculate the number of representative particles in 2.15 moles of gold.<br>

LekaFEV [45]

Answer:

$1.29 * 10^{24}$ particles of gold

Explanation:

To convert the number of moles of any substance, in this case gold, you need Avogadro's number.

Avogadro's number is always $6.022$ × $10^{23}$

$2.15$ moles Au × $\frac{6.022*10^{23} particles}{1 mole Au}$ = $1.29 * 10^{24}$ particles of gold

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3 years ago

Joel has noticed that his engine is not compressing properly during the compression and ignition step. What problem will this po

denis-greek [22]

The answer is: The engine will run inefficiently APEX....

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3 years ago

What experimental evidence was provided for the nuclear model of an atom

UkoKoshka [18]

Alpha particles bouncing off of gold foil.

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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

What are some examples of minerals that occur as elements instead of<br>compounds

katrin [286]

Answer: Aluminum, Antimony, Arsenic, Bismuth, Carbon, Cadmium, Chromium, Cobalt, etc.

3 0

3 years ago