sos, may day! help

[NE2000]

n(Ca(OH)2) = c x V = 0.35 x 0.300=0.105 mol
n(OH-) = 2 x 0.105 = 0.21 mol

n(H+) = 0.12 mol

OH- in excess = 0.09 mol
Total volume = 0.6L
c(OH-) = 0.15 M
pOH = 0.82
Therefore pH=14-0.82=13.18
Therefore answer is D

Alternatively you can avoid the pOH thing and use [H+][OH-] = 10^-14.

[TrueTears]

First write equation.



mol

mol

Ratio

We have . So for 0.105 mol of we need mol of HCl . But we have , so therefore n(HCl) is limiting.

so we only need mol of . Therefore left over is mol







therefore



Hence D

[lacoste]

thanks!!!!!!



Which of the following reactions is NOT a redox process?
A 2K(s) + Br2(l) →2KBr(s)
B Ca(OH)2(s) + CO2(g) →CaCO3(s) + H2O(l)
C NH4NO3(s) → N2O(g) + 2H2O(g)
D CH4(g) + Cl2(g) →CH3Cl(g) + HCl(g)

[chem-nerd]

B is the only one where oxidation numbers haven't changed

[lacoste]

What is the oxidation numbers of A?

also, how do you know which reductant goes to which oxidant etc?

thanks .

[Gloamglozer]

For these sort of questions, I just use process of elimination first and then I use oxidation numbers.

It cannot be A because it has an element (K).  All equations containing an element is redox.

I'd say B because the oxidation numbers haven't changed.

EDIT:  Damn, beaten to it.  lol

[lacoste]

Can monosaccharides be hyrdolysed to form smaller units?

??

[TrueTears]

Can monosaccharides be hyrdolysed to form smaller units?

??

no.

[Mao]

Can monosaccharides be hyrdolysed to form smaller units?

??

not into smaller units, but it can be hydrolysed into a straight chain, which is slightly heavier that the ringed structure (extra H2O).

[lacoste]

If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.

Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)

[hyperblade01]

If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.

Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)

Looks like a limiting/excess reactant question but actually this one is double excess :p

Using the equation it reads:

1 Mole Fe2O3 and 3 Mole CO   gives    2 Mole Fe and 3 Mole CO2

Therefore if 1 mole of Fe is produced, you divide everything by 2:

0.5 Mole Fe2O3 and 1.5 Mole CO   gives    1 Mole Fe and 1.5 Mole CO2

3 - 0.5 = 2.5 Mole Fe2O3 remaining
2 - 1.5 = 0.5 Mole Co remaining

[lacoste]

Thats what I did to work it out but the solution is 5.5 mol, is there a different way to work it out?

Also, another q, the pH of a 10^-9M solution of HCl is closest to :

A 4
B 6
C 7
D 9


thanks.

[Mao]

a 10^-9 M HCl would completely ionise and form 10^-9 M of [H+]. However, water at neutral point has [H+] of 10^-7. Hence addition of 10^-9 M HCl makes neutral water 1% more acidic, pH is still 7.

[lacoste]

a 10^-9 M HCl would completely ionise and form 10^-9 M of [H+]. However, water at neutral point has [H+] of 10^-7. Hence addition of 10^-9 M HCl makes neutral water 1% more acidic, pH is still 7.

man, thats confusing. is this likely to be on the exam?

If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.

Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)

Why is the answer 5.5 mol? Is this wrong?

Thanks.

[chem-nerd]

If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.

Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)

Looks like a limiting/excess reactant question but actually this one is double excess :p

Using the equation it reads:

1 Mole Fe2O3 and 3 Mole CO   gives    2 Mole Fe and 3 Mole CO2

Therefore if 1 mole of Fe is produced, you divide everything by 2:

0.5 Mole Fe2O3 and 1.5 Mole CO   gives    1 Mole Fe and 1.5 Mole CO2

3 - 0.5 = 2.5 Mole Fe2O3 remaining
2 - 1.5 = 0.5 Mole Co remaining

Thats what I did to work it out but the solution is 5.5 mol, is there a different way to work it out?

amount of substance at the end of the reaction would be from the reactants AND products. so you have to add the 1 mol of Fe and 1.5 mol of CO₂ produced