Virgil Said:

Physical Science help would be lovely. Care to assist me?

We Answered:

I just wanted to congratulate you on using the Homework Help section in, what I would think is the appropriate and logical way to get help. I can't actually help you with the question, but I'm glad you at least took a shot at what you thought the answer was rather than asking us to do your homework #1- 20 for you. :)

Good Luck (I'm gonna agree with B also though! )

Marc Said:

chemistry help!!!!!!!!!!!!! with bronsted lowry and lewis acid/bases?

We Answered:

According to bronsted-lowry an acid is a proton donor and a base is a proton acceptor
An example of an acid is HCl because it dissociates and produces H+ ions in water.
____

The two that donate hydrogen ions are HNO3 and H3O+
because HNO3-->H+ + NO3-
and H3O+-->H2O + H+
______

There are two other types of acids and bases:

Lewis acid: accepts electron pairs
Lewis base: donates electron pairs

Arrhenius acid: donates H+
Arrhenius base: donates OH- (hydroxide ion)

Knowing that you should be able to answer any of those questions.

Terry Said:

Acid/Base Worksheet-Please Help!?

We Answered:

pX= - log[X] so:
pH= - log [H+]
pOH= - log [OH-]

pOH + pH =14
[OH-][H+]=10^-14

10^-pH = {H+]
a- LiOH acts as a base (because it has OH) so pOH=- log [0.000669]
and then pH=14 - (- log [0.000669])
or you could solve [OH-][H+]=10^-14 where OH- is 0.000669

b- 10^-1.18=[H+] so you have c and v=3.44L so use n=cv

c-pOH=-log0.771

With these basic steps you should be able to solve the rest but if you can't I advise you post them again and split them up

James Said:

Acids/Bases (Practice worksheet) Neep help?

We Answered:

4.Write the "two" dissociations for H2SO4
Answer:
a)H2SO4 + H2O = H30= + HSO4-
b)HSO4- + H2O = H30+ + SO4 2-

5. Three different models of acids and bases (I think its wrong)
Answer:
a) Arrhenius Theory
b) Bronsted/Lowry Theory
c) Lewis Theory

8. What color would the chemicals be if tested with phenolphthalein?
Answer:
a)HCl - No Change
b)NaOH - pink
c)HCN - No Change
d)KOH - pink
e)NH3 - Pink
f)HC2H3O2 - No Change

Cecil Said:

Chemistry Help on Acids and Bases?

We Answered:

1) The volume of NaOH used = final buret - initial buret
Trial 1 - 44.72 ml - 3.55 ml = 41.17 ml
Trial 2 = 43.79 ml - 2.80 ml = 40.99 ml
Trial 3 = 45.20 ml - 4.05 ml = 41.15 ml
Average volume NaOH used = 41.10 ml

2) Molarity = moles / Litres
Therefore moles = molarity x L
So moles NaOH = conc NaOH x Average volume (in L)
moles NaOH = 0.500 M x 0.04110 L
= 0.02055 moles NaOH were needed to neutralise the vinegar
= 0.0206 moles (3 sig figs)

3.)
The balanced equation is
CH3COOH + NaOH ------> CH3COONa + H2O

You can see that 1 mole of CH3COOH needs 1 mole of NaOH to be neutralised

So, since 0.0206 moles of NaOH was needed there must have been 0.0206 moles of acetic acid in the vinegar