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
545mm Hg in Kilopascals is 72.6607
I hope this helps you. Good luck stay safe, healthy and, happy!<3
Answer:
K = 137.55 atm/M.
Explanation:
- The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:
<em>P = (K)(C)</em>
where P is the partial pressure of the gaseous solute above the solution (P = 1.0 atm).
k is a constant (Henry’s constant).
C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).
∴ K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.
Explanation:
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Answer:
![Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OH%5D%7D)
Explanation:
Hello,
In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:
Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:
![K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}](https://tex.z-dn.net/?f=K%3DKa%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OH%5D%7D)
Best regards.
