Chemistry: Post your doubts here!

[Snowysangel]

And in this too:
Which element of the 3rd period requires the least number if moles of oxygen for the complete combustion of 1 mil of the element?
A) aluminum
B) magnesium
C) phosphorus
D) sodium

 

[daredevil]

HELP!! What happens when you react sodium chloride with nitric acid?

i googled it just to be sure...
it makes sodium nitrate and hydrochloric acid.

 

[daredevil]

And in this too:
Which element of the 3rd period requires the least number if moles of oxygen for the complete combustion of 1 mil of the element?
A) aluminum
B) magnesium
C) phosphorus
D) sodium

3/2O2 + Al -->Al2O3

Mg + O --> MgO

P + 5/4O2 ---> 1/4P4O10

Na + 1/2O2 --> 1/2Na2O


so i guess sodium is the answer

 

[Snowysangel]

i googled it just to be sure...
it makes sodium nitrate and hydrochloric acid.

But sodium nitrate is soluble.. Could you explain this question instead?

 

[Snowysangel]

3/2O2 + Al -->Al2O3

Mg + O --> MgO

P + 5/4O2 ---> 1/4P4O10

Na + 1/2O2 --> 1/2Na2O


so i guess sodium is the answer

Thanks! But shouldn't it be: Mg + 0.5 O2 --> MgO?

 

[Young Stunner]

How do we do this?

 

[sidbloom1995]

its q. 3d
Can you also solve that question of pka of on2006 when you knew the year

i was able to understand it but i wasnt able to solve it (the third part) ~sorry; was not much help~
(i) they asked us the pH, for that we will dissociate the acid and form an equation
CH3CH2COOH -----------> CH3CH2COO- + H+
we use this equation in in the formula Ka=[CH3CH2COO-]x[H+]/[CH3CH2COOH]
we have the conc. for [CH3CH2COOH] = 0.5 and Ka = 1.34 x 10^-5
since we have only one H+ and one rest of the molecule CH3CH2COO-, we can assume that they will have equal concentrations! so we will take them as "x"
equating everything:

1.34 x 10^-5 = [x][x]/[0.5]
1.34 x 10^-5 x [0.5] = [x]^2
take square root on both sides and you will get
[x] = 2.6 x 10^-3

now use this conc. in the equation pH= -log[H+]
you will get 2.6 pH

(ii) our equation will be:
NaOH + CH3CH2COOH ---------> [CH3CH2COO-Na+] + H2O
1 : 1 ----------> 1 : 1
0.03 : 0.05 ----------> 0.03 : -
~this part is not asked, it's related to the next part~

(iii) they first found out the no. of moles of the acid, by using it's concentration and volume, we got, 0.5 x 0.1 = 0.05 mol
according to our equation the mole ratio between them is of [1:1] and they already gave us the moles of NaOH as 0.03 mol
0.05 was the no. of moles present in our solution and the no. that reacted with the NaOH was 0.03 (since our acid:base was 1:1) so we subtract them, 0.05-0.03 = 0.02 mol
0.02 mol is the amount of acid left and the 0.03 mol is used up to make sodium propanoate
in this question the concept of excess moles is applied!! the acid has to be in excess or it will all be used up and neutralized, base on the other hand will be all used up will be 100% converted to salt, sodium prppanoate, thus our no. of moles for NaOH = no. of moles of sodium proponoate
thus our sodium proponate is 0.03 mol ~do you get it or is it starting to get confusing?~

(iv) in this part we use two equations, pKa = -logKa -----> eq.1 and the equation for the pH of buffer
pH = pKa + log([salt]/[acid])------> eq.2.................||in case of a base buffer the equation changes pH = (Kw - pKb) + log([base]/[salt]), ours is an acid buffer so we dont need this one right now||
use the eq.1 to find pKa = -log[ 1.34 x 10^-5] = 4.87,
equate it in eq.2
pH = 4.87 + log([0.03]/[0.02]) = 5.046

that will be all
as for that previous qustion i solved it just recently so i'll try to look for it again and tell you when i find the ans! inshallah!

 

[sitooon]

i was able to understand it but i wasnt able to solve it (the third part) ~sorry; was not much help~
(i) they asked us the pH, for that we will dissociate the acid and form an equation
CH3CH2COOH -----------> CH3CH2COO- + H+
we use this equation in in the formula Ka=[CH3CH2COO-]x[H+]/[CH3CH2COOH]
we have the conc. for [CH3CH2COOH] = 0.5 and Ka = 1.34 x 10^-5
since we have only one H+ and one rest of the molecule CH3CH2COO-, we can assume that they will have equal concentrations! so we will take them as "x"
equating everything:

1.34 x 10^-5 = [x][x]/[0.5]
1.34 x 10^-5 x [0.5] = [x]^2
take square root on both sides and you will get
[x] = 2.6 x 10^-3

now use this conc. in the equation pH= -log[H+]
you will get 2.6 pH

(ii) our equation will be:
NaOH + CH3CH2COOH ---------> [CH3CH2COO-Na+] + H2O
1 : 1 ----------> 1 : 1
0.03 : 0.05 ----------> 0.03 : -
~this part is not asked, it's related to the next part~

(iii) they first found out the no. of moles of the acid, by using it's concentration and volume, we got, 0.5 x 0.1 = 0.05 mol
according to our equation the mole ratio between them is of [1:1] and they already gave us the moles of NaOH as 0.03 mol
0.05 was the no. of moles present in our solution and the no. that reacted with the NaOH was 0.03 (since our acid:base was 1:1) so we subtract them, 0.05-0.03 = 0.02 mol
0.02 mol is the amount of acid left and the 0.03 mol is used up to make sodium propanoate
in this question the concept of excess moles is applied!! the acid has to be in excess or it will all be used up and neutralized, base on the other hand will be all used up will be 100% converted to salt, sodium prppanoate, thus our no. of moles for NaOH = no. of moles of sodium proponoate
thus our sodium proponate is 0.03 mol ~do you get it or is it starting to get confusing?~

(iv) in this part we use two equations, pKa = -logKa -----> eq.1 and the equation for the pH of buffer
pH = pKa + log([salt]/[acid])------> eq.2.................||in case of a base buffer the equation changes pH = (Kw - pKb) + log([base]/[salt]), ours is an acid buffer so we dont need this one right now||
use the eq.1 to find pKa = -log[ 1.34 x 10^-5] = 4.87,
equate it in eq.2
pH = 4.87 + log([0.03]/[0.02]) = 5.046

that will be all
as for that previous qustion i solved it just recently so i'll try to look for it again and tell you when i find the ans! inshallah!

May allah shine you with all A*'s for your help to others
A little more explaination of how NaOH and the salt concentration are equal will be appreciated

 

[sidbloom1995]

May allah shine you with all A*'s for your help to others
A little more explaination of how NaOH and the salt concentration are equal will be appreciated

ok
check it in the pictures!

 

[sitooon]

ok
check it in the pictures!

Did you try the other question ?

 

[sidbloom1995]

Did you try the other question ?

which one?

 

[sidbloom1995]

i checked your question again!
are you asking about the smooth curves?
just as the names suggest they are curves that are smooth, we usually use then in reaction kinetics most often!
was this what you were asking?

 

[sitooon]

i checked your question again!
are you asking about the smooth curves?
just as the names suggest they are curves that are smooth, we usually use then in reaction kinetics most often!
was this what you were asking?

The question from on2006 about pka

 

[sidbloom1995]

The question from on2006 about pka

oh ok!
that questuion was asked by someone else and a specific part so i didnt do the pKa part before!
(i) the Ka expression
Ka = [H+][R-]/

---

---

---

............the R is the rest of the group or structure!....in the MS they wrote A but that was confusing because the previous question asked for a compound name A, so i used R instead! ( R is more commonly used to represent groups or structure that are not involved!)

(ii) the conc. of Ibuprofen is 0.15 mol/dm3 and Ka = 6.3 x 10^-6
since ibuprofen have only one carboxylic acid group, (unlike A which have 2)
we know that they divide to form 1 [H+] and 1 [R-], so we know that they have equal conc. and we can donate both of the as "x"

6.3 x 10^-6 = [x][x]/[0.15]
[x]^2 = [6.3 x 10^-6] x [0.15]
take square root on both sides
x = 9.72 x 10^-4
now use the formula pH= -log[H+]
pH= -log[9.72 x 10^-4]
pH= 3.01

 

[sitooon]

oh ok!
that questuion was asked by someone else and a specific part so i didnt do the pKa part before!
(i) the Ka expression
Ka = [H+][R-]/

---

---

---

............the R is the rest of the group or structure!....in the MS they wrote A but that was confusing because the previous question asked for a compound name A, so i used R instead! ( R is more commonly used to represent groups or structure that are not involved!)

(ii) the conc. of Ibuprofen is 0.15 mol/dm3 and Ka = 6.3 x 10^-6
since ibuprofen have only one carboxylic acid group, (unlike A which have 2)
we know that they divide to form 1 [H+] and 1 [R-], so we know that they have equal conc. and we can donate both of the as "x"
6.3 x 10^-6 = [x][x]/[0.15]
[x]^2 = [6.3 x 10^-6] x [0.15]
take square root on both sides
x = 9.72 x 10^-4
now use the formula pH= -log[H+]
pH= -log[9.72 x 10^-4]
pH= 3.01

---

I am very sorry , i meant the LAST part !!

 

[midha.ch]

Someone please help me with Inorganic Chemistry! I need good easy to understand notes!!
I would really be hankful if someone helps me! Need it asap

 

[sitooon]

Someone please help me with Inorganic Chemistry! I need good easy to understand notes!!
I would really be hankful if someone helps me! Need it asap

 

[sitooon]

Someone please help me with Inorganic Chemistry! I need good easy to understand notes!!
I would really be hankful if someone helps me! Need it asap

 

[sitooon]

Someone please help me with Inorganic Chemistry! I need good easy to understand notes!!
I would really be hankful if someone helps me! Need it asap

I hope i could help you more ,
you are the only one to answer my doubts here !

 

[sitooon]

Someone please help me with Inorganic Chemistry! I need good easy to understand notes!!
I would really be hankful if someone helps me! Need it asap

And this is the summary of ALL reactions!