Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
The substance has a higher density than water
JJ Thompson discovered electrons
Answer:
The concentration of protons affects an enzyme's folded structure and reactivity.
Explanation:
Enzymes act within narrow pH limits (optimal reaction pH). Since most enzymes have a protein structure, the variation in pH or temperature affects their enzymatic activity.
To catalyze a reaction, an enzyme binds to one or more reagent molecules. These molecules are the substrates of the enzyme.
In some reactions, a substrate breaks into several products. In others, two substrates join together to create a larger molecule or to exchange parts. In fact, for any biological reaction that can occur to you, there is probably an enzyme to accelerate it.
The part of the enzyme where the substrate binds is called the active site.
The amino acid residues of the active site often have acidic or basic properties that are important for catalysis. Changes in pH can affect these residues and make binding with the substrate difficult.
The answer is na3po4+3koh=3naoh+k3po4
