reactions to break down glucose using oxygen to produce carbon dioxide, water and energy in the form of ATP. ... To balance the oxygen atoms for the reactant side, you need to count 6 atoms from the glucose.
Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

And the resulting percent yield:

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Answer:
For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions
For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14
So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) = 1*10^-9 OH ions
as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.
If you overheat copper sulfate higher of mass will be lost that is copper sulfate will loss sulfur and oxygen which led to a higher loss of mass than if you would have heated enough. This higher mass lost will be shown in calculation as percentage of water lost