Question

i need help doing a lab. the name is Carbon Dioxide from Antacid Tablets. unit 1 please help.

Answer 1

@BellaBlueBird

Answer 2

im sorry but i cant help ya on this one i havent learned that yet sowwy

Answer 3

thanks for trying @BellaBlueBird

Answer 4

we can apply this formula, in order to compute the requested volume:
\[\Large V = \frac{{{C^3}}}{{6{\pi ^2}}}\]
where \(C\) is the circumference

Answer 5

can you help me set it up please? @Michele_Laino

Answer 6

I can help with chemical reactions and stoichiometry

Answer 7

first step:
the third column \(Volume\), can be completed using my formula above and the data from the second column of such table

Answer 8

for example, the first row, of the third column, contains this value:
\[\Large V = \frac{{{{\left( {20.74} \right)}^3}}}{{6 \cdot {{\left( {3.14} \right)}^2}}} \simeq 150.8\;c{m^3}\]

Answer 9

@Ilovepurple42043

Answer 10

@Michele_Laino i got the volume of each of them... but now im stuck on the questions. can you help me?

Answer 11

I think that in the subsequent chemical reaction:
\[\Large {\text{NaHC}}{{\text{O}}_3} + {\text{ }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{C}}{{\text{O}}_{\text{2}}} + {\text{ NaOH }} + {\text{ }}{{\text{H}}_{\text{2}}}{\text{O}}\]
we have the decoposition of sodium carbonate \({\text{NaHC}}{{\text{O}}_3}\)

Answer 12

how do u know? and can you help with the other 2? @Michele_Laino

Answer 13

as we can see, at the rigt side of the equation of reaction, we have an aqueous solution of sodium carbonate, now, such salt will dissociate in \(Na^+\) ions and \({\text{HC}}{{\text{O}}_3^-}\) negative ions. Furthermore, the \({\text{HC}}{{\text{O}}_3^-}\) ions will decompose in such way:
\[{\text{HCO}}_3^ - \rightleftarrows {{\text{H}}^ + } + {\text{CO}}_3^ - \]

Answer 14

oops.. I meant:
\[\Large {\text{HCO}}_3^ - \rightleftarrows {{\text{H}}^ + } + {\text{CO}}_3^{2 - }\]

Answer 15

oops.. I meant \(left\) side not right side

Answer 16

question #3
hint:
the molar mass of \({\text{NaHC}}{{\text{O}}_3}\) is: \(84.01\), so the mol of \({\text{NaHC}}{{\text{O}}_3}\), are:
\[\Large \frac{2}{{84.01}} = ...?\]
From the stoichiometry of the reaction, we see that we get:
\[\Large \frac{2}{{84.01}} = ...?\]
mol of \(CO_2\), so, the requested quantity of \(CO_2\), is:
\[\Large \frac{2}{{84.01}} \cdot 44.01 = ...?\]
since the molar mass of \(CO_2\) is \(44.01\)