Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch.
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
Answer:-
People with diabetes must monitor their blood glucose levels constantly.
Blood glucose levels are nowadays measured by the help of enzyme glucose oxidase. The enzyme glucose oxidase catalyzes the oxidation of glucose to gluconic acid.
By this way all of the glucose is measured at one time. At the same time hydrogen peroxide is produced.
The hydrogen peroxide reacts with a second color producing chemical. The concentration of the glucose can be related to the intensity of color produced
True, because the large covalent molecules would combine with the others