Answer:
Potassium iodide increases the decomposition rate of hydrogen peroxide.
Explanation:
Potassium iodide increases the decomposition rate of hydrogen peroxide because potassium iodide act as a catalyst. A catalyst speed up the process of chemical reaction without reacting with the molecules present in reaction. If the potassium iodide is not present as a catalyst for the decomposition of hydrogen peroxide then the decomposition of hydrogen peroxide takes too much time because the catalyst is absent that speed up the reaction.
Answer:
1.4 × 10^-4.
Explanation:
C3H6O3 + H2O <======> C3H5O3^- + H3O^+ ------------------------------------------(1).
So, from the question above we are given the following parameters or data which is going to help in solving this particular Question/problem;
=>concentration of the solution of lactic acid (CH3CH(OH)C00H) = 0.1 M and pH = 2.44.
Therefore, the concentration of the hydrogen ion[H^+} can be determined from the pH formula given below;
pH = - log { H^+}.
2.44 = - log { H^+}.
Therefore, {H^+} = 0.0036 M.
From the equation (1) given above, we have that the ratio for the equilibrium reaction is 1 : 1 : 1 :1. Therefore, molarity of C3H5O3^- = 0.0036 M and the molarity of C3H6O3 =( 0.1 - 0.0036 M) = 0.0964 M at equilibrium.
Hence, ka = {C3H5O3^-} { H3O^+} /{C3H6O3} = ( 0.0036 M)^2 /(0.0964 M) = 1.4 × 10^-4.
If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.
<h3>What is a strong acid?</h3>
A weak acid is one that is able to ionize completely in solution. The acid called chromic acid H2CrO4 is not able to ionize completely in solution.
We know that a buffer is composed of a weak acid and its salt or a weak base and its salt hence if the acid H2CrO4 is present in a buffer then the other half must be salt of the acid.
If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.
Learn more about buffer:brainly.com/question/22821585
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<span>Solutes become electrolytes by ionizing. Ionic compounds therefore make good electrolytes; covalent compounds don't.</span>
Explanation:
a) when zinc burnt in oxygen.
2Zn + O2 -----∆-----> 2ZnO(black residue)
b) when carbon burnt in oxygen.
C+O2----∆---> CO2.
c) when sulphur burnt in oxygen.
S+O2-----∆-----> SO2.
d) when Calcium burnt in oxygen.
2Ca+O2-----∆-----> 2CaO(black residue)
e) when Magnesium burnt in oxygen.
2Mg+O2-----∆----> 2MgO.
f) when sodium burnt in oxygen.
4Na+O2----∆-----> 2Na2O.
hope all these reactions help you.
