Answer:
2HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)
Explanation:
HCl(aq) + Na2SO3(aq) —> NaCl(aq) + H2SO3(aq)
Let us balance the equation. This is illustrated below:
There are 2 atoms of Na on the left side of the equation and 1atom on the right side. It can be balance by putting 2 in front of NaCl as shown below:
HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)
Now, we have 2 atoms of Cl on the right side and 1 atom on the left side. Thus, it can be balance by putting 2 in front of HCl as shown below:
2HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)
A careful look at the equation proved that the equation is balanced as the numbers of the different atoms of the element on both side of the equation are the same.
Answer: All of the statements are true.
Explanation:
(a) Considering the system mentioned in the equation:-
The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B
(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.
0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,
40 to 60s is 0.14 (0.30-.16 = 0.14) mol
(0.46 > 0.14) mol leading this statement to be true as well.
(c) Average rate from t1 = 40 to t2 = 60 s is given by:
which is true as well
HBr reacts with LiOH and forms LiBr and H₂O as the products. The balanced reaction is
LiOH(aq) + HBr(aq) → LiBr(aq) + H₂O(l)
Molarity (M) = moles of solute (mol) / volume of the solution (L)
Molarity of LiOH = 0.205 M
Volume of LiOH = 29.15 mL = 29.15 x 10⁻³ L
Hence,
moles of LiOH = molarity x volume of the solution
= 0.205 M x 29.15 x 10⁻³ L
= 5.97575 x 10⁻³ mol
The stoichiometric ratio between LiOH and HBr is 1 : 1.
Hence,
moles of HBr in 25.0 mL = moles of LiOH added
= 5.97575 x 10⁻³ mol
Hence, molarity of HBr = 5.97575 x 10⁻³ mol / 25.00 x 10⁻³ L
= 0.23903 M
≈ 0.239 M
Hence, the molarity of the HBr is 0.239 M.
We are asked for the ratio of ions to produce neutral KCl. When neutral potassium chloride dissociates, the reaction is KCl = K+ + Cl-. Hence, the ratio of ions from the dissociation is 1 mole potassium ion per mole chlorine ion or 1 mole chlorine ion per potassium ion.
The moles of potassium that you would need to prepare 1200 g of 5% potassium sulfate solution is 1.538 moles
calculation
calculate the mass potassium using the below formula
%M/M = mass of the solute(potassium)/mass of the solvent (potassium sulfate solution)
let the mass of potassium be represented by Y
then convert % into fraction = 5/100
5/100 = Y/1200
cross multiplication
100y = 6000
divide both side by 100
Y= 60 g
moles of potassium =mass/molar mass
= 60/39=1.538
