Answer:
41.16 moles of H2O
Explanation:
Ratio for the products-reactants is 1:6, so 1 mol of glucose is produced when plants use 6 moles of water.
Then, let's make a rule of three:
1 mol of glucose is produce by using 6 moles of water
6.86 moles of glucose are produced by the use of (6 . 6.86)/1 = 41.16 moles of H2O
if 1 g is equal to 100 cg
then 0.55 g are equal to X cg
X = (0.55 × 100 ) / 1 = 55 cg
The density of the object is 55 cg/L.
Answer:

Explanation:
Hello,
In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

Clearly, it turns out negative since the concentration is diminishing due to its consumption.
Regards.
Answer: C excess reactant
Reason: It is not catalyst because catalyst are not consumed. It is not product because it was one of the reactants. It is not limiting because there was excess left over.