Answer:
The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM
Explanation:
The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.
Here's that idea written as a formula: c= n/V
where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.
You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L
Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L
Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,

, the reaction will be

and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4
And
![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
In this experiment, the experimental group is ice, the independent variable is sunlight while on the other hand, the dependent variable is block of ice.
<h3>What are dependent and independent variables?</h3>
Dependent variable is a type of variable in which a variable whose value depends on another variable while on the other hand, independent variable is a type of variable in which a variable whose value does not depends on another variable. Dependent variable is denoted by X whereas the independent variable is denoted by Y.
So we can conclude that In this experiment, the experimental group is ice, the independent variable is sunlight while on the other hand, the dependent variable is block of ice.
Learn more about variable here: brainly.com/question/25223322
#SPJ1