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Leona [35]
1 year ago
12

calculate no. of molecules in a spherical drop of water... radius = 0.1 mm and density of water = 1g/cm²​

Chemistry
1 answer:
SVEN [57.7K]1 year ago
8 0

Answer:

1.49X10^20

Explanation:

hope this helps (´・ω・`)

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Federico has two samples of pure water—sample X and sample Y. Sample X has a volume of 1 L, and sample Y has a volume of 10 L. H
Lostsunrise [7]

Answer:

The boiling point of sample X and sample Y are exactly the same.

Explanation:

The difference between sample X and sample Y is that they occupy different volumes. However, they both contain pure water. Remember that pure water has uniform composition irrespective of its volume.

Volume does not affect the boiling point as long as the volume is small enough not to give rise to significant pressure changes in the liquid.

The boiling point of a liquid is the temperature at which the pressure exerted by the surroundings upon a liquid is equaled by the pressure exerted by the vapour of the liquid; under this condition, addition of heat results in the transformation of the liquid into its vapour without raising the temperature.

It can be clearly seen from the above that the volume of a solution of pure water does not affect its boiling point hence sample X and sample Y will have the same boiling point.

8 0
3 years ago
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
3 years ago
Pls help me, ill mark brainliest and give 5 stars
Maslowich

Answer:

Decomposition!

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The answer is Photosphere Apex.      

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I think the answer is
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