An experiment produced 0.10 g $CO_2$, with a volume of 0.056 L at STP. If the accepted density of $CO_2$ at STP is 1.96 g/L, what is the approximate percent error?

The percent error is basically the fractional deviation from the accepted value. The percent error is the absolute difference between the measured/approximate and accepted/exact values, relative to the exact value:...

The formula for percent error is: % error = $(x_(measured)-x_(true))/x_(true)*100%$ Our $x_(true)$ value is 5.5 m. All we do now is solve for % error by plugging in each result....

The formula for percent error is Percent error$=abs((E-T)/T) * 100$ where E = experimental value and T= theoretical or accepted value Percent error $=abs((2.04-2.00)/2.00) * 100$ $=abs(.04/2.00) * 100...

$CO_2$ taken in by the plant initially bonds to ribulose 1,5-biphosphate or $RuBP$ to form an unstable 6-carbon compound. This then divides into two three-carbon of glycerate 3-phosphate. With the...

$"Moles of methane"$ $=$ $(5.563*g)/(16.043*g*mol^-1)$ $=$ $0.3468*mol$. $"Moles of dioxygen"$ $=$ $(25.00*g)/(31.998*g*mol^-1)$ $=$ $0.7813*mol$. Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation: $CH_4(g) +...

There are actually two methods of doing it - stoichimetry & also with the help of law of conservation of mass. First by : $16$grams of $CH_4$ means $1$mole of...

We use the Ideal Gas equation to access the moles of carbon dioxide evolved: $n=(PV)/(RT)$ $=$ $((750*mm*Hg)/(760*mm*Hg*atm^-1)xx75.0xx10^-3*L)/(0.0821*L*atm*K^-1*mol^-1xx293*K)$ $=3.1xx10^-3mol$. $"Mass of carbon dioxide produced"=3.1xx10^-3molxx44.0*g*mol^-1=??g$ The key to doing this problem was...

use the formula $mols=(mass)/(mr)$ mr is the relative . so mols are unknown mr of CO2: $12+16(2)$ $44$ mass is $12.6g$ now put them in the formula given $mols=12.6/44$...

By definition there are $6.022xx10^23$ carbon dioxide molecules in a $44.0*g$ mass. How do I know this? Well because we use the given number to specify molar mass, the which...

.......where $N_A="Avogadro's Number"$ $=$ $6.022xx10^23*mol^-1$. And so, $1.70xx10^-2*molxxN_A=1.70xx10^-2*cancel(mol)xx6.022xx10^23*cancel(mol^-1)$, approx. $1.0xx10^22$ $"carbon dioxide molecules"$. We get the answer as a number as required.