Rishi's Chemistry Thread

[Rod]

Stupid question, but where did this come from? C6H8O7

That's the formula for citric acid.

You gave me the formila for glucose (c6h12o6), so I ignored it because your question wanted citric acid.

[Rishi97]

That's the formula for citric acid.

You gave me the formila for glucose (c6h12o6), so I ignored it because your question wanted citric acid.

OMG!!! That is soo embarassing...
Thanks heaps for that Rod 

[alchemy]

Hey Rishi

Assuming it's this reaction 3NaOH + C6H8O7 → NaH2C6H5O7 +3 H2O

n(NaOH) = 0.350 X 0.03709 = 0.01298 MOL
n (citric acid) = 1/3 x 0.00432 mol
c (Citric acid) + 0.00432 /0.02 = 0.216 mols per litre
c (citric acid) = 0.216 x 192 = 41.5
= C
Assuming it's the reaction I have given you.

I think you made some mistakes on the parts I coloured in red. Your final answer is correct though, you just typed some of your steps in wrong I guess.

Firstly, the equation should be  C6H8O7 + 3NaOH --> Na3C6H5O7 + 3H2O. Note how the sodium gets the 3 from the citric acid when it citric acid loses/donates it's 3 protons.
Also, you typed in the second line of your working incorrectly. It should read 1/3 * 0.01298. The rest of your steps are correct (:

[Rod]

OMG!!! That is soo embarassing...
Thanks heaps for that Rod 

No it's not . And no probs

I think you made some mistakes on the parts I coloured in red. Your final answer is correct though, you just typed some of your steps in wrong I guess.

Firstly, the equation should be  C6H8O7 + 3NaOH --> Na3C6H5O7 + 3H2O. Note how the sodium gets the 3 from the citric acid when it citric acid loses/donates it's 3 protons.
Also, you typed in the second line of your working incorrectly. It should read 1/3 * 0.01298. The rest of your steps are correct (:

Thanks but didn't really make any difference, I used 0.01298 mol when I multiplied by a third

mehs all this for a multi choice question xD

[alchemy]

mehs all this for a multi choice question xD

I know it was an MCQ, but I just wanted to provide Rishi with the steps she would need to apply if asked that in a short answer question.

[Rishi97]

I know it was an MCQ, but I just wanted to provide Rishi with the steps she would need to apply if asked that in a short answer question.

Much appreciated alchemy
Thanks

[alchemy]

Much appreciated alchemy
Thanks

No problem (:

[Rishi97]

What is the concentration in mol/L, of I^- (aq) in a solution that is 5.00% KI, by mass, if the solution has a density of 1.038 g/ml?
a) 0.0301
b) 0.0313
c) 0.313
d) 0.625

Thanks

[nhmn0301]

What is the concentration in mol/L, of I^- (aq) in a solution that is 5.00% KI, by mass, if the solution has a density of 1.038 g/ml?
a) 0.0301
b) 0.0313
c) 0.313
d) 0.625

Thanks

5.00% of KI meaning 5g of KI in 100g of solution.
Density of solution = m/V
1.038= 100/V
V = 96.3 mL. Hence, there are 5g of KI in 96.3mL solution.
n (KI) = m/M = 5 / 166 = 0.0301 (mol) = n(I-)
Hence, there are 0.0301 mol of I- in 96.3mL
                   =>  0.0301 mol in 0.0963 L
                    =>  0.313 mol in 1L
So I guess the answer is C. Let me know if there is any errors!

[Rishi97]

Consider the following equation:

2H₂O₂(l) --> 2H₂O (l) + O₂ (g)
In this reaction:
a) H₂O₂ is undergoing oxidation only
b) H₂O₂ is undergoing reduction only
c) H₂O₂ is undergoing both oxidation and reduction
d) H₂O₂ is not undergoing either reduction or oxidation

[nhmn0301]

Consider the following equation:

2H₂O₂(l) --> 2H₂O (l) + O₂ (g)
In this reaction:
a) H₂O₂ is undergoing oxidation only
b) H₂O₂ is undergoing reduction only
c) H₂O₂ is undergoing both oxidation and reduction
d) H₂O₂ is not undergoing either reduction or oxidation

Normally, when you assign oxidation number, you let O = -2. But in here, we have H2O2, so the order to the rules becomes important, if you have a look back to the order, the rule which states that H is a +1 is above O is a -2. Hence, in H2O2, we assign H as +1 and O as -1. Hence, you have O goes from -1 -> -2, which means it's being reduced. So I reckon it's B.

[Rishi97]

Normally, when you assign oxidation number, you let O = -2. But in here, we have H2O2, so the order to the rules becomes important, if you have a look back to the order, the rule which states that H is a +1 is above O is a -2. Hence, in H2O2, we assign H as +1 and O as -1. Hence, you have O goes from -1 -> -2, which means it's being reduced. So I reckon it's B.

The answer says D

[nhmn0301]

The answer says D

Sorry about that, my bad.  But even if I need to choose again, I would choose C. The half equation I think goes like this:
H2O2 -> O2 + 2H+ + 2e^- oxidation
2e- + 2H+ + H2O2 -> 2H2O reduction
That's what I think but according to the question, it's wrong so please correct me.

[lzxnl]

Question must be wrong then. You've given the equation for the spontaneous (albeit slow if uncatalysed) decomposition of hydrogen peroxide, which is interesting as it's a disproportionation reaction, when H2O2 is both oxidised and reduced simultaneously. The answers given must be wrong. O2 has oxidation state zero for oxygen, while H2O has oxidation state -2 and H2O2 has oxidation state -1

In weird cases like these, don't just go by rules. Draw out the structure, which is H-O-O-H
Each hydrogen is bonded to an oxygen => oxidation state convention says each hydrogen has 'lost' an electron as oxygen is more electronegative and has oxidation state +1. Then, each oxygen has only gained one electron each from the hydrogen as the O-O bond is completely symmetric => oxygen gains one electron, oxidation state -1. No need for 'dodgy' application of rules

[Rishi97]

When methane reacts with chlorine in the presence of uv light,
a) an addition reaction occurs to form CH₃Cl
b) a substitution reaction occurs to form CH₃Cl
c) the products are CH₃Cl and HCl only
d) the products include CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄ and HCl