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2 years ago

7

Why isn’t a state change from liquid to solid a chemical reaction?

2 answers:

kipiarov [429]2 years ago

4 0

Answer:

because the change can be undone with a chemical change the change cant be undone, and it just passes through another phase of matter. Changes of matter can consider physical

Explanation:

DanielleElmas [232]2 years ago

3 0

Answer:

nosé ehhhye eee p nosé inglés

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In the lewis structure of ch3oh, how many lone pairs of electrons are there? in the lewis structure of ch3oh, how many lone pair

Lunna [17]

The structure of Methanol (CH₃OH) is shown below. This structure contains 2 lone pair of electrons on oxygen (highlighted red). Electronic configuration of oxygen is,

Oxygen = 8 = 1s², 2s², 2px², 2py¹, 2pz¹

There are six electrons in valence shell of oxygen, among these six electrons two electrons are unpaired and involved in forming covalent bond with hydrogen and carbon, while remaining 4 electrons in pair (i.e. 2s², 2px²) exists as lone pairs.

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3 years ago

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3 years ago

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mrs_skeptik [129]

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16

Explanation:

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3 years ago

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3 years ago

Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9

EleoNora [17]

Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

q = charge of sodium ion

n = number of sodium ion = $2.68\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

4 0

3 years ago