The zinc content of a 1.50 g ore sample was determined by dissolving the ore in HCI, which reacts with the zinc. The excess HCI is then neutralized with with NaOH. The reaction of HCI with Zn is shown. Zn(s) + 2 HCI(aq) - ZNCI, (aq) + H,(g) The ore was dissolved in 150 mL of 0.600 M HCI, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 9.95 mL of 0.536 M NAOH to neutralize the excess HCI. What is the mass percentage (%w/w) of Zn in the ore sample? %w/w = % Zn etv AA A O O hulu MacBook Pro

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Chapter15: Acid-base Equilibria
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The zinc content of a 1.50 g ore sample was determined by dissolving the ore in HCl, which reacts with the zinc. The excess HC.
is then neutralized with with NaOH. The reaction of HCl with Zn is shown.
Zn(s) + 2 HCl(aq) → ZnCl, (aq) + H, (g)
The ore was dissolved in 150 mL of 0.600 M HCI, and the resulting solution was diluted to a total volume of 300 mL. A
20.0 mL aliquot of the final solution required 9.95 mL of 0.536 M NAOH to neutralize the excess HCl. What is the mass
percentage (%w/w) of Zn in the ore sample?
%w/w =
% Zn
O étv AA A OO
O hulu
MacBook Pro
Co
G Search or type URL
%
&
24
4.
8.
9
delete
Y.
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OP
MOSISO
command
option
Transcribed Image Text:The zinc content of a 1.50 g ore sample was determined by dissolving the ore in HCl, which reacts with the zinc. The excess HC. is then neutralized with with NaOH. The reaction of HCl with Zn is shown. Zn(s) + 2 HCl(aq) → ZnCl, (aq) + H, (g) The ore was dissolved in 150 mL of 0.600 M HCI, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 9.95 mL of 0.536 M NAOH to neutralize the excess HCl. What is the mass percentage (%w/w) of Zn in the ore sample? %w/w = % Zn O étv AA A OO O hulu MacBook Pro Co G Search or type URL % & 24 4. 8. 9 delete Y. U OP MOSISO command option
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