how many liters of nitrogen gas are needed to react with 90.0 g of potassium at STP in order to produce potassium nitride accord
2 answers:
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The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>First, let's look at the equation of the reaction:
The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
More on percent yield can be found here: brainly.com/question/17042787
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Answer :
The concentration of before any titrant added to our starting material is 0.200 M.
The pH based on this ion concentration is 0.698
Explanation :
First we have to calculate the concentration of before any titrant is added to our starting material.
As we are given:
Concentration of HBr = 0.200 M
As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion and bromide ion
.
As, 1 M of HBr dissociates to give 1 M of
So, 0.200 M of HBr dissociates to give 0.200 M of
Thus, the concentration of before any titrant added to our starting material is 0.200 M.
Now we have to calculate the pH based on this ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
Thus, the pH based on this ion concentration is 0.698