how do different stimuli affect the survival behavior of an organism?THIS IS THE SECOND TIME IM ASKING PLZZZZ HELP ME
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Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = moles = 0.00600 moles
Neutralization reaction (back titration):
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%