I believe the answer is D
Answer:
C the mass of each product formed
Explanation:
To the determine the limiting reactant, it is essential that we have the balanced equation of the reaction from which we can calculate the stochiometry mole ratio of the reactant. After this, we need to calculate the molar mass of the reactants, using the mole from the balanced equation we can calculate each mass of each reactant needed. Finally we need the mass of each reactant using proportion we can calculate the amount needed for the reaction from the masses of the reactant by comparing the mass given against the mass calculated from the balanced equation. After this, the mass that is exhausted or that is finished will be the limiting reactant which is the reactant that finished and caused the reaction to stop.
Answer:
The answer to your question is Molarity = 0.41
Explanation:
Data
mass of KNO₃ = 76.6 g
volume = 1.84 l
density = 1.05 g/ml
Process
1.- Calculate the molecular mass of KNO₃
molecular mass = 39 + 14 + (16 x 3) = 101 g
2.- Calculate the number of moles
101 g of KNO₃ --------------- 1 mol
76.6 g of KNO₃ ------------ x
x = (76.6 x 1) / 101
x = 0.76 moles
3.- Calculate molarity
Molarity = 
Substitution
Molarity = 
Result
Molarity = 0.41
<span> The body has to do this to make sure and or to prevent damage to the delicate tissues that form the Respiratory System. This is important because </span>Your body's organs, such as your heart and brain, need oxygen-rich blood to work well. (if it gets damaged by the Cold/Dry air you breath in on a daily)<span><span>Respiratory failure </span>also can occur if
your lungs can't properly remove carbon dioxide (a waste gas) from your
blood. Too much carbon dioxide in your blood can harm your body's
organs.
Hopefully i helped ^.^ please mark brainly if its a possibility~
</span>
<h2 /><h2 /><h2>answer.</h2>
five oxygen molecules
step by step explanation.
according to the equation,one molecule of oxygen is enough to react with two carbon molecules thus 10 carbon molecules need 5oxygen molecules
