In the following reaction, 48.26g of sodium permanganate, NaMnO4, is reacted with 59.50g of sodium sulfite. Assuming an excess of water, which reactant is the limiting reagent and how many grams of the excess reactant is left at the end of the reaction? 2NaMnO4 + 3Na2SO3 + H2O → 2MnO2 + 3Na2SO4 + 2NaOH

Just took the quiz on edgenuity and the correct answer is "c.  4.54 x  10–12  j"

Answerstrong acids and bases completely dissociate in water.this is how strong acids and bases are defined. they will ionize 100% in solution, making a non-reversible reaction

First up write the aim
then write the materials reqd.
then the procedure which you follow
then observation
then conclusion
then you can write lab safety rules or precautions(you can avoid if u want).
lemme' give an example
aim: to observe the double displacement reaction
materials reqd: bacl2 and na2so4, test tubes and a dropper
procedure:
1. take bacl2 in a test tube and na2so4 in the other.
2. then pour bacl2 into na2so4 solution.
3. observe
observation:
you will see that a white precipitate is formed called baso4. along with that common salt or nacl is formed.
the equation is
bacl2+na2so4=baso4+2nacl
conclusion:
a double displacement reaction is observed.
hope it

Try adding 700% because of the x in y