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2 years ago

Convert 3.4 x 10^23 molecules of NaCl to grams

Chemistry

1 answer:

shusha [124]2 years ago

7 0

Answer:

$m_{NaCl}=33g$

Explanation:

Hello,

In this case, in order to compute the grams of sodium chloride starting by the molecules, the first step is to compute the moles contained in the given amount of molecules by using the Avogadro's number:

$n_{NaCl}=3.4x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} =0.56mol$

Then, by using the molar mass of sodium chloride (58.45 g/mol) we can directly compute the grams:

$m_{NaCl}=0.56mol*\frac{58.45g}{1mol} \\\\m_{NaCl}=33g$

Regards.

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frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

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(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

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Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

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Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

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