Answer:
42.64 M
Explanation:
Step 1: Given data
- Moles of HCl (solute): 8.1 mol
- Volume of the solution: 189.98 mL
Step 2: Convert the volume of the solution to liters
We will use the relationship 1 L = 1,000 mL.

Step 3: Calculate the molarity of the solution
The molarity is equal to the moles of solute divided by the liters of solution.

<h3>
Answer:</h3>
1170.43 m³
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial pressure, P1 as 808 kPa
- Initial temperature, T1 as 585 K
- Initial volume, V1 as 295 m³
- New pressure, P2 as 102 kPa
- New temperature, T2 as 293 K
We are required to find the new volume;
- We are going to use the combined gas law
- According to the gas law;

- Thus, rearranging the formula;



Therefore, the volume is 1170.43 m³
The half cell in which the electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.
<h3><u>What is a Galvanic cell ?</u></h3>
Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.
Two half-cells make up most electrochemical cells. The half-cells allow electricity to pass via an external wire by separating the oxidation half-reaction from the reduction half-reaction.
<h3><u>
Oxidation:</u></h3>
The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.
<h3><u>Reduction:</u></h3>
The second half-cell, cathode, which is frequently displayed on a figure's right side. The cathode is where reduction happens. The circuit is completed and current can flow by adding a salt bridge.
To know more about processes in Galvanic cell, refer to:
brainly.com/question/13031093
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Answer:
The molar mass in g/mol is 121.4 g/m
Explanation:
Let's apply the Ideal Gases Law to solve this:
P . V = n . R. T
V = 125 mL → 0.125L
P = 754 Torr
760 Torr ___ 1 atm
754 Torr ____ (754 / 760) = 0.992 atm
Moles = Mass / Molar mass
0.992 atm . 0.125L = (0.495 g / MM) . 0.082 . 371K
(0.992 atm . 0.125L) / (0.082 . 371K) = (0.495 g / MM)
4.07x10⁻³ mol = 0.495 g / MM
MM = 0.495 g / 4.07x10⁻³ mol → 121.4 g/m