Asked by Karen
How many millimoles of HCl (aq) are neutralized by 0.35 g of CaCO3 (sol)? mmol
Answers
Answered by
bobpursley
it takes two moles of this acid to neutralize one mole of chalk.
millimolesHCl=.35/molmassCaCO3 * 2molHCL/1molCaCO3 *1mol/1000millimol
= .35/100*2*1E-3 check that.
millimolesHCl=.35/molmassCaCO3 * 2molHCL/1molCaCO3 *1mol/1000millimol
= .35/100*2*1E-3 check that.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.