Question #01c42

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Question #01c42

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Answer 1 · 2018-01-09T04:58:39

$C a t i o n s$ $Cations$ = $H^{+}$ $H^+$ $i o n s$ $ions$ = $1.204 \times 10^{24} H^{+}$ $1.204xx10^24 H^+$

Explanation:

$Given . mass . of$ $"Given"color(white)(.) "mass"color(white)(.) "of"$ $H_{2} S O_{4}$ $H_2SO_4$ = $196 . g r a m s$ $196color(white)(.)grams$
=> $moles . of, H_{2} S O_{4}$ $"moles"color(white)(.) "of"color(white)(,) H_2SO_4$ = $\frac{196 g}{98 \frac{g}{mole}}$ $(196g)/(98g/"mole"$ = $2.13 .$ $2.13color(white)(.)$ $. moles$ $color(white)(.)"moles"$ $H_{2} S O_{4}$ $H_2SO_4$

Assuming 100% ionization:
=> $1 . mole . H_{2} S O_{4}$ $1color(white)(.)"mole"color(white)(.) H_2SO_4$ => $2 . moles . i o n s H^{+}$ $2color(white)(.)"moles" color(white)(.)ions H^+$ + $1 . mole . S O_{4}^{2 -} . ions$ $1color(white)(.)"mole" color(white)(.)SO_4^(2-)color(white)(.)"ions"$

=> $2.13 . moles . H_{2} S O_{4}$ $2.13color(white)(.) "moles"color(white)(.) H_2SO_4$ => $4.26 .$ $4.26"color(white)(.)$ moles $. H^{+} i o n s$ $color(white)(.)H^+ions$
=> $4.26$ $4.26$ $.$ $color(white)(.)$ $cancel("mole"H^+ion")$ $xx$ $(6.023xx10^23$ $H^+ioncolor(white)(.)"particles"$ / $1.00color(white)(.)cancel("mole"H^+ions)$ )

=> $1.204xx10^24$ $H^+ions$

NOTE: The $H^+$ ion is also the 'Hydronium Ion' $H_3O^+$

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Question #01c42

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