Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
Answer:
Explanation:
General reaction of acid in water is as follows:
HCl + H2O = H3O+ + Cl-
Thus Acids increase the concentration of hydronium ions in solution by donating hydrogen ions to water molecules is true
NH₃, being a basic gas neutralizes the HNO₃ forming a salt NH₄NO₃
Therefore the correct answer is NH₃ and NH₄NO₃
The solution of which only 32% dissociates to release OH⁻ ions is a weak base. This is because some of the energy is used when the substance reacts with the solution thus some bonds are not broken.
HCl is an acid. This is because it dissociates in water to give H⁺ as the only positively charged ions.
Arrhenius acid increases the concentration of hydrogen ions because it dissociates to release hydrogen ions as the only positively charged ions in the acid. So the answer is TRUE
Arrhenius base dissociates in water to release hydroxide ions as the only negatively charged ions.
NaOH⁺aq⇒Na⁺ ₍aq₎+ OH⁻₍aq₎
<u>Answer:</u> The value of 
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
Given mass of 
= 9.2 g
Molar mass of = 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:
For the given chemical equation:
<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of = 0.057
Evaluating the value of 'x'
The expression of for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:
Hence, the value of for the given reaction is 1.435
