Mario places 10 mL of water in a test tube and heats the liquid over a Bunsen burner for 2 minutes. After removing the test tube from the Bunsen burner, there are 6 mL of water left in the test tube. This experiment is a good example of a <span>physical change involving phase changes. </span>
In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half-cell. A similar (but reversed) situation is found in the cathodic cell.
<h3>
What purpose does a salt bridge serve in an oxidation process?</h3>
Anions (negatively charged particles) are added to the solution of the oxidation half of the cell by the salt bridge, and cations (positively charged particles) are added to the solution of the reduction half of the reaction.
<h3>
What purpose does the salt bridge serve in a galvanic cell?</h3>
For instance, KCl, AgNO3, etc. In a galvanic cell, such as a voltaic cell or Daniel cell, salt bridges are typically used. A salt bridge's primary job is to assist in preserving the electrical neutrality of the internal circuit. Additionally, it aids in keeping the cell's response from reaching equilibrium.
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- According to Pascal’s principle, for a certain fluid in a totally enclosed system, a change in pressure at a given point in the fluid is transferred to all points in the fluid, as well as to the enclosing walls.
- This is illustrated by the fact that the pressure inside an enclosed system is the same according to the relation [ pressure = force/area ]. Therefore, the change in pressure resulting from squeezing a ketchup bottle will be transferred equally to all parts of that bottle as well as its internal content.
- That’s how hydraulic machines, such as garbage trucks and hydraulic lifts function..
We are given the equation to use which is:
ΔG = ΔH - TΔS
We are also given that:
ΔG = 173.3 kJ
T = 303 degrees kelvin
ΔH = 180.7 kJ
Substitute with these givens in the above equation to get ΔS as follows:
ΔG = ΔH - TΔS
173.3 = 180.7 - 303ΔS
303ΔS = 180.7 - 173.3
303ΔS = 7.4
ΔS = 7.4 / 303 = 0.02442 kJ/K which is equivalent to 24.42 J/k
Based on the above calculations, the correct choice is:
D. 24.42 J/K
Answer:
sub-particle charge mass
protons +1 1
neutron 0 1
electron - 1 negligible
protons and neutrons are found in the nucleus
electrons in the shells orbiting the nucleus
