POH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁹ ]
pOH = 9
Answer C
hope this helps!
Answer: 7.07 grams
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.052 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 1 mole of
Thus 0.052 moles of give = of
Mass of
Thus 7.07 g of will be produced from the given masses of both reactants.
Look on this website!
http://lawr.ucdavis.edu/classes/ssc100/CEC_Answers02.htm