We are given the chemical reaction and the amount of reactant used for the process. We use these data together to obtain what is asked. We do as as follows:
0.882 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 0.441 mol O2 produced
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the semipermeable membrane surrounding the cytoplasm of a cell.
Answer:
70.0 %
Explanation:
Step 1: Given data
- Mass of nitrogen (mN): 74.66 g
- Mass of the compound (mNxOy): 250 g
Step 2: Calculate the mass of oxygen (mO) in the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
mNxOy = mN + mO
mO = mNxOy - mN
mO = 250 g - 74.66 g = 175 g
Step 3: Determine the percent composition of oxygen in the sample
We will use the following expression.
%O = mO / mNxOy × 100%
%O = 175 g / 250 g × 100% = 70.0 %
The significant figure is only 1 which is just the number 5. The zeros don't count.
Answer:
B. NO
Explanation:
In each mole of NO, the weight of the oxygen is 16.00 g; the weight of the nitrogen is 14.01 g. Then in 80 g of NO, we have 42.65 g of oxygen and 37.35 g of nitrogen.
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In each mole of NO₂, the weight of the oxygen is 2(16.00) = 32 g. The weight of the nitrogen is 1(14.01) = 14.01 g.
From the available oxygen, we can produce ...
(42.65 g +16 g)/(32.00 g/mol) = 1.83 mol of NO₂
From the available nitrogen, we can produce ...
(37.35 g)/(14.01 g/mol) = 2.67 mol of NO₂
Clearly, the reaction is limited by the amount of available oxygen.
NO is the excess reactant.
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