20LL 2.Write a net ionic equation for the reaction of acetic acid with sodium bicarbonate, if the bicarbonate is already dissolved in water when the reaction occurs.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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2.6. The red colour in cabbage is due to the presence of a range of anthocyanins such as
cyanidin. Draw both the molecular structure for cyanidin (showing all carbon and
hydrogen atoms) and the line structure (using lines and not showing carbon atoms or
hydrogen atoms that are attached to carbon atoms). You can search online to find the
correct structure for cyanidin.
Transcribed Image Text:2.6. The red colour in cabbage is due to the presence of a range of anthocyanins such as cyanidin. Draw both the molecular structure for cyanidin (showing all carbon and hydrogen atoms) and the line structure (using lines and not showing carbon atoms or hydrogen atoms that are attached to carbon atoms). You can search online to find the correct structure for cyanidin.
CH3CoONa present in
aqueous state ,will be present
as ions CH3 CO
CH3 00H is also prsent in
aqueou'stale, in ions it be
CHACO
S) + H eq) + NaHCO3 (s) nd
>CH3C00 cag)+Ne'caj) + H20
2.Write a net ionic equation for the reaction of acetic acid with sodium bicarbonate, if the
bicarbonate is already dissolved in water when the reaction occurs.
13 How many moles of sodium bicarbonate are present in the 5.00 g you used for this
experiment?
メ3
mass of Na - 23, H= 1, C=12:0=16
HC0335.00g
い ャる +1 = 84gmo)
2+16×3
%3D
Transcribed Image Text:CH3CoONa present in aqueous state ,will be present as ions CH3 CO CH3 00H is also prsent in aqueou'stale, in ions it be CHACO S) + H eq) + NaHCO3 (s) nd >CH3C00 cag)+Ne'caj) + H20 2.Write a net ionic equation for the reaction of acetic acid with sodium bicarbonate, if the bicarbonate is already dissolved in water when the reaction occurs. 13 How many moles of sodium bicarbonate are present in the 5.00 g you used for this experiment? メ3 mass of Na - 23, H= 1, C=12:0=16 HC0335.00g い ャる +1 = 84gmo) 2+16×3 %3D
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