Answer:
B. 111 J
Explanation:
The change in internal energy is the sum of the heat absorbed and the work done on the system:
ΔU = Q + W
At constant pressure, work is:
W = P ΔV
Given:
P = 0.5 atm = 50662.5 Pa
ΔV = 4 L − 2L = 2 L = 0.002 m³
Plugging in:
W = (50662.5 Pa) (0.002 m³)
W = 101.325 J
Therefore:
ΔU = 10 J + 101.325 J
ΔU = 111.325 J
Rounded to three significant figures, the change in internal energy is 111 J.
Answer: C) the values of Kb and Kw
Explanation: i just took the test
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
Lead (II) chloride+ potassium nitrate
PbCI2+KNO3(B)
