Oluwasaeon:
It's simple na

number5:


Bro pls help

number5:
Hello guys.. I need urgent help from any chemistry guru in the house to help solve this.. Just got home very tired and my younger sister is crying on me to do this assignment for her... Save a brother's sister from cane pls, google hasn't been of good help here

P. S just solve and snap and paste here pls.. Very urgent

dejt4u:

1a; standardize the solution to get the concentration of B in g/dm³
i.e (2.65g/250cm³)×1000cm³ = 8.60g/dm³

1b; use the formula
[(C_A × V_A)/(C_B × V_B)] = N_A/N_B,
Where C_A and C_B rep. Conc. of A and B respectively. V_A and V_B rep. conc. of A and B respectively. N_A/N_B is the mole ratio of A to B

In this case, A is the acid, (the HCl) and B is the Base (Na₂CO₃)

C_A is what you are asked to calculate,
C_B is your answer in 1a above divided by the Molar mass of B,
V_A is 20cm³, V_B is 25cm³, N_A = 2 and N_B = 1

number5:


Is there a need to convert the 25cm3 and 20cm3 to dm3 by dividing by 1000?

dejt4u:
not necessary,.. Even if you convert them, you're still going to get the same answer as the units will eventually cancel each other.. e.g 1÷2 is the same thing as 1000÷2000

I've inserted the equation for the reaction

dejt4u:
For Q2
i; Conc. of B in mol/dm³ can be calculated using the formula [(C_A × V_A)/(C_B × V_B)] = N_A/N_B,

So C_B = (C_A × V_A × N_B)/(V_B × N_B),
C_A = 0.16M (or 0.16mol/dm³)
V_A = 13.20cm³
V_B = 25.00cm³
N_A = 2 and
N_B = 1

number5:


Tnx bro.. I'm following keenly

dejt4u:
I've updated my previous post for Q2(iii)

number5:


Tnx bro.. I'm copying it as you're updating

dejt4u:
done with Q2.. Check it and let me know if there is a part you don't understand

number5:


Tnx bro..

In d last question under number 2, i. Substitutin for the atomic number of each element, we get something like na2Co3xH20 = 23*2 +12+16*3 + x + 1*2 +16

=106+ x + 18... Im lost here, how do i deal with the X

dejt4u:
Done with Q3

number5:


Still left with one bro

.there was no space so i wrote it at d top of the book
C. Molar mass of the dibasic acid

dejt4u:
OK.. Since we are not sure of the Dibasic acid, we can't determine it's molar mass directly by adding the Atomic masses.. But we can get it's molar mass by using the formula,

Molar mass = (Mass conc. g/mol) / (Molar Conc in mol/dm³)

Molar conc is our answer to Q3(ii) above
Mass conc is (3.5g/250cm³) × 1000cm³ which is equal to 14.0g/dm³

number5:


Ok.. Wow.. Tnx so much bro.. For all d efforts, i deeply appreciate

dejt4u:
you're very welcome. This is the aspect of chemistry I enjoyed most while in secondary school. Thank you for taking me back the memory lane

dejt4u:
you're very welcome. This is the aspect of chemistry I enjoyed most while in secondary school. Thank you for taking me back the memory lane

number5:
tnx bro.. Also been long i did dis aspects of chemistry

chidiebereuzoma:

Please we are waiting for you at nnpc thread.
Something that requires your attention as the history keeper of our great house popped up

dejt4u:
For Q2
i; 2HCl + Na₂CO₃ --> 2NaCl + H₂O + CO₂

ii; Conc. of B in mol/dm³ can be calculated using the formula [(C_A × V_A)/(C_B × V_B)] = N_A/N_B,

So C_B = (C_A × V_A × N_B)/(V_B × N_B),
C_A = 0.16M (or 0.16mol/dm³)
V_A = 13.20cm³
V_B = 25.00cm³
N_A = 2 and
N_B = 1

iii; molar mass of B = (mass conc in g/dm³)/(Conc of B in mol/dm³)
To get mass conc in g/dm³, standardize the solution, i.e (3g/250cm³) × 1000cm³ = 12g/dm³

Also, Conc of B in mol/dm³ is the answer to Q2(ii) above

iv; To get the value of X, sub for mass numbers of Na, C, H and O and equate everything to the molar mass calculated in Q(iii) above

i.e Na₂CO₃ + XH₂O = Answer in Q(iii),
23(2) + 12 + 16(3) + X((1×2)+16) = Answer in Q(iii),
106 + 18X = Answer in Q(iii),
Then collect the like terms and make X the subject of formula
(That is your final answer)


v; % by mass of water of crystallization = (Molar Mass of XH₂O) divided by Molar Mass of B (i.e Na₂CO₃.XH₂O) × 100%

NB: Molar Mass of Na₂CO₃.XH₂O, is the answer in Q2 (iii).
To get molar mass of XH₂O, substitute for the atomic mass of each elements in the compound and the value of X we got in Q2(iv)

number5:


Lolz.. Doing elder brother duties for now.. Be there soon

dejt4u:
I made a slight mistake in Q2(iv) and Q2(v) and I've corrected it.. Kindly check the updated version. Sorry for that

number5:


Yes. I noticed and corrected it last night.. Tnx