OK SO todayyyy we looked at page 7 in the YELLOW booklet thing.
Page 7:
What volume of a 0.250 ml/L solution of carbonic acid is needed to neutralize 30.0g of sodium hydroxide?
Q1: write the nutralization reaction
A1: H2CO3 (aq) + 2NaOH -> Na2CO3(aq) + 2H2O(s)
Q2: Determine the number of moles of base.
A2: 30g NaOH x (1mol NaOH/40g NaOH) = .75 mol NaOH
Q3: Calculate moles of H2CO3 Needed to neutralize the base.
A3: 1mol acid 2 mol base
.375 mol acid .75 mol base
Q4: Calculate the volume of H2CO3 required
A4: V=n/c = (.375 mol H2CO3)/(.250 mol H2CO3)
V=1.5 H2CO3(aq)
WE ALSO GOT a NEw booklet on Neutralization Reactions and did the questions on the back of the booklet :D
It was overall a productive day andd personally, i don't understand what we covered ... But its not too too bad .
OH!!! AND we had a lock down in effect and ppl were yelling i the halls and so on so that was kinda the most interesting peek of the class :)
NEXT SCRIBE IS ***NCC-74656***
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1 comment:
Wow, such large letters. Good review of the class & happy birthday in advance.
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