$CO_2$ 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...
1 Answers 1 viewsThere 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...
1 Answers 1 viewsBy 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...
1 Answers 1 viewsYou have $3.90xx10^23$ molecules, and the molar quantity is the simply the quotient, $(3.90xx10^23*"molecules")/(6.022xx10^23*"molecules"*mol^-1)$ $=$ $??mol$
1 Answers 1 viewsThe balanced chemical equation tells us unequivocally that 16 g of methane gas reacts with 32 g of dioxygen gas to give 44 g of carbon dioxide, and 36 g...
1 Answers 1 viewsHere $CH_4$ being excess $O_2 $ will be consumed fully. We know from above equation 2 mole or 64 g $O_2$ produces 1 mole or 44 g $CO_2$ So 125...
1 Answers 1 viewsThis is a limiting reactant problem. We know that we will need a balanced equation and moles of each reactant. 1. Gather all the information in one place with...
1 Answers 1 views$10gramsCaCO_3(s) => CaO(s) + (?)O_2(g)$ $"moles" CaCO_3 = ((10g)/(100(g/(mol))))$ = $0.10"mole"$ From reaction ratios => moles $CaCO_3(s)$ consumed = moles $O_2(g)$ produced = 0.10 mole $(O_2(g))$**** If a 'dry'...
1 Answers 1 viewsThe idea here is that you need to figure out how many moles of carbon dioxide, $"CO"_2$, are initially dissolved in the soda by using the volume and of...
1 Answers 1 viewsMass is a measure of the number of particles. Here you have $O$ atoms, and dioxygen, $O_2$, and ozone molecules, $O_3$ but each have the same mass by specification. There...
1 Answers 1 views