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podryga [215]
3 years ago
9

Some type of path is necessary to join both half-cells in order for electron flow to occur.

Chemistry
1 answer:
klemol [59]3 years ago
4 0

The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

Explanation:

Flow of electrons is possible with the help of a conducting medium like metal wire.

A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.

If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.

Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

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What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?
Vladimir79 [104]
Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2  -------> 32 g O2 -----------> 36 g H2O
   ↓                       ↓                             ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.


4 0
3 years ago
Name three examples of how the atom has changed since Rutherford's discovery of the nucleus .
NikAS [45]

Answer:

Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment, in which he demonstrated that the atom has a tiny, high- mass nucleus. In his experiment, Rutherford observed that many alpha particles were deflected at small angles while others were reflected back to the alpha source.

8 0
3 years ago
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sp2606 [1]

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3 years ago
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the
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Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

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=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol

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