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nirvana33 [79]
3 years ago
12

The forces between water molecules are stronger than the forces between ethanol molecules. Which liquid would probably be most d

ifficult for an insect to walk on? Explain your answer.
Chemistry
2 answers:
valentinak56 [21]3 years ago
6 0

Answer:

An insect would have an easier time walking on the surface of water than on the surface of ethanol.

Water’s stronger intermolecular forces lead to higher surface tension.

Higher surface tension allows water to support the insect.

Explanation:

WITCHER [35]3 years ago
4 0
<span>An insect would have an easier time walking on the surface of water than on the surface of ethanol. Water's stronger intermolecular forces lead to higher surface tension. Higher surface tension allows water to support the insect. I hope this helps.</span>
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Which law relates to the ideal gas law?
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<u>Answer:</u> The law that related the ideal gas law is \frac{V_1}{n_1}=\frac{V_2}{n_2}

<u>Explanation:</u>

There are 4 laws of gases:

  • <u>Boyle's Law:</u> This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

Mathematically,

P_1V_1=P_2V_2

  • <u>Charles' Law:</u> This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

  • <u>Gay-Lussac Law:</u> This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

  • <u>Avogadro's Law:</u> This law states that volume is directly proportional to number of moles at constant temperature and pressure.

Mathematically,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

Hence, the law that related the ideal gas law is \frac{V_1}{n_1}=\frac{V_2}{n_2}

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Why sometimes the electrons of an atom transfer to higher energy level​
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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
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<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

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}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

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n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

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