Answer:
Ca(NO3)2 has the highest boiling point ( option A)
Explanation:
Step 1: Data given
A. 1.25 M Ca(NO3)2
B. 1.25 M KNO3
C. 1.25 M CH3OH
D. 2.50 M C6H12O6
Step 2: Calculate highest boiling point
The boiling point depends on the van't Hoff factor
This shows the particles produced when the substance is dissolved. For non-electrolytes dissolved in water, the van' t Hoff factor is 1.
Ca(NO3)2 → Ca^2+ + 2NO3- → Van't Hoff factor = 3
KNO3 → K+ + NO3- → Van't Hoff factor = 2
CH3OH is a non-elektrolyte → Van't Hoff factor = 1
C6H12O6 is a non-elektrolyte → Van't Hoff factor = 1
Ca(NO3)2 has the highest boiling point
A cone of depression occurs in an aquifer when groundwater is pumped from a well. In an unconfined aquifer, this is an actual depression of the water levels. In confined aquifers, the cone of depression is a reduction in the pressure head surrounding the pumped well.
Answer:
9 : 8
Explanation:
Aluminum oxide has the following formula Al₂O₃.
Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:
Mass of Al in Al₂O₃ = 2 × 27 = 54 g
Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g
Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:
Mass of Al = 54 g
Mass of O₂ = 48 g
Mass of Al : Mass of O₂ = 54 : 48
Mass of Al : Mass of O₂ = 54 / 48
Mass of Al : Mass of O₂ = 9 / 8
Mass of Al : Mass of O₂ = 9 : 8
Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8
Answer:
The molarity of the formed CaBr2 solution is 0.48 M
Explanation:
Step 1: Data given
Number of moles CaBr2 = 0.72 moles
Volume of water = 1.50 L
Step 2: Calculate the molarity of the solution
Molarity of CaBr2 solution = moles CaBr2 / volume water
Molarity of CaBr2 solution = 0.72 moles / 1.50 L
Molarity of CaBr2 solution = 0.48 mol / = 0.48 M
The molarity of the formed CaBr2 solution is 0.48 M
Answer:
Explanation:
1. Calculate the initial moles of acid and base
2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
