The first step is to find the number of moles of OH⁻ that reacted with the HCl. To do this multiply 2.00L by 1.50M to get 3 moles of Ca(OH)₂. Then you multiply 3 by 2 (there are 2 moles of OH⁻ per every 1 mole of Ca(OH)₂) to get 6 moles of OH⁻. That means that you needed 6 moles of HCl since 1 mole of HCl contains 1 mole of H⁺ and equal amounts H⁺ and OH⁻ reacted with each other. To find the molarity of the HCl solution you need to divide 6mol by 1L to get 6M. Tat means that the concentration of the acid was 6M.
I hope this helps. Let me know if anything was unclear.
Only one. variable to get a reliable results you only have to change one.
Answer:
The final temperature was 612 °C
Explanation:
Charles's law relates the volume and temperature of a certain amount of ideal gas, maintained at a constant pressure, using a constant of direct proportionality. In this law, Charles says that at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

When you want to study two different states, an initial and a final one of a gas and evaluate the change in volume as a function of temperature or vice versa, you can use the expression:

In this case:
- V1= 5.76 L
- T1= 22 °C= 295 °K (Being 0°C=273°K)
- V2=17.28 L
- T2=?
Replacing:

Solving:

T2= 885 °K = 612 °C
<u><em>The final temperature was 612 °C</em></u>
Answer:
1.The electrode on the right is positive
2. 0.058V
Explanation:
The above cell is a concentration cell.
A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).
Part 2: Please, see the attachment below for the calculations.