Answer:
Explanation:
K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl
.033 M .053 M
Ksp of Ba CrO₄ is 2.10×10⁻¹⁰
Ksp of ( COO ) ₂ Ba is 1.30×10⁻⁶
A ) Ksp of Ba CrO₄ is less so it will precipitate out first .
B) Ksp = 2.10×10⁻¹⁰
Ba CrO₄ = Ba⁺² + CrO₄⁻²
C .033
C x .033 = 2.10×10⁻¹⁰
C = 63.63 x 10⁻¹⁰ M
Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M
C)
90% of precipitation of barium oxalate
concentration of oxalate to precipitate out = .9 x .0532 = .04788
( COO ) ₂ Ba = (COO)₂⁻² + Ba⁺²
.04788 M C
C x .04788 = 1.30×10⁻⁶
C = 27.15 x 10⁻⁶ M .
Answer:
Explanation:
The mass of radioactive isotope at a certain time is given by the following expression:
Time constant can be calculed in terms of half-life:
The initial mass is:
Answer:
a. Concave down
Linear increasing
b. Increases the reaction rate
c. The reaction approaches the saturation point of the enzyme
Explanation:
a. For the reaction with enzyme, the shape is concave down. The action of the enzyme on the preferred substrate is initially very rapid and decreases as the enzyme becomes saturated and the ratio of products to substrate increases to approach an equilibrium rate of reaction
For the reaction without enzyme, the shape is linear and increasing. Increase in the concentration of the substrate will increase the number of effective collisions that lead into product formation leading to an increased rate of the chemical reaction
b. The enzyme increases the proportion of effective combination of substrates to form the products
c. The curve of the reaction with enzyme flattens out because as the concentration of the substrate increases while that of the enzyme remains the same, the enzyme becomes saturated and less able to increase the rate of the reaction of the excess substrate.
So what I know is that enzyme and substrate are like lock and key meaning that when the active site of the enzyme changes, the enzyme will not fit to the substrate which will lead the enzyme to denature. Hope this helps.
Molarity is a measure of a solution's concentration calculation by getting the ratio of the number of moles of solute to the total volume of solution. This has a unit of M or molar, equivalent to mole/L.
It is more important and meaningful to know the molarity rather than if the solution is dilute or concentrated because molarity gives the QUANTITATIVE approach of knowing the concentration while the second one only gives us the QUALITATIVE description of the solution. Hence, we are able to calculate for other unknown parameters if we have the molarity known.
