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3 years ago

Which statement is true of diffusion?

Chemistry

1 answer:

raketka [301]3 years ago

4 0

Answer:

Molecules move from areas of high concentration to areas of low concentration.

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Can someone pls help, I’m sorry if it’s a lot

Helga [31]

Answer: 2738

Explaining: you gotta add

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3 years ago

When something is in the incorrect to state that a compound is broken down into its component elements in a decomposition reacti

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3 years ago

Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

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

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

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:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

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3 years ago

If a scientist accidentally spills a chemical that causes invalid conclusions, he has made__________________?

iren [92.7K]

It is B) An honest mistake

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3 years ago

Read 2 more answers

a given sample of a gas has a volume of 3.0 L at a pressure of 4.0 atm. If the temperature remains constant and the pressure is

pentagon [3]

Answer:-

2.0 L

Explanation:-

Initial Volume = V 1 = 3.0 L

Initial Pressure P 1 = 4.0 atm

New pressure P 2 = 6.0 atm

Using boyle's law

P1 V1 = P2 V2

New Volume V2 = P1 V1 / P2

= 4.0 atm x 3.0 L / 6.0 atm

= 2 .0 L

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3 years ago