Across a period I.E increases progressively from left to right
Explanation:
The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.
- Ionization energy is a measure of the readiness of an atom to lose an electron.
- The lower the value, the easier it is for an atom to lose an electron.
- Elements in group I tend to lose their electrons more readily whereas the halogens hold most tightly to them.
- The first ionization energy is the energy needed to remove the most loosely bonded electron of an atom in the gaseous phase.
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Answer:
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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.
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Answer:
0.1 M
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 100 mL
Initial concentration (C1) = 0.5 M
Final volume (V2) = 500 mL
Final concentration (C2) =?
Using the dilution formula C1V1 = C2V2, the new concentration of the solution can be obtained as follow:
C1V1 = C2V2
0.5 × 100 = C2 × 500
50 = C2 × 500
Divide both side by 500
C2 = 50/500
C2 = 0.1 M
Therefore, the new concentration of the solution is 0.1 M