Answer:
22:
Formular:

substitute:

23:
<em>Same</em><em> </em><em>element</em><em> </em><em>is</em><em> </em><em>represented</em><em> </em><em>by</em><em> </em><em>same</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>protons</em><em>.</em><em> </em>
Answer:
6 protons. 6 protons
7 neutrons. 8 neutrons
6 electrons. 6 electrons
Note: <u>Atoms</u><u> </u><u>with</u><u> </u><u>same</u><u> </u><u>proton</u><u> </u><u>number</u><u> </u><u>but</u><u> </u><u>different</u><u> </u><u>mass</u><u> </u><u>number</u><u> </u><u>are</u><u> </u><u>called</u><u> </u><u>isotopes</u>
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
The correct answer is D :)
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
M1V1 = M2V2
1% x V1 = 0.25% x 10 mL
V1 =2.5 mL
Therefore, you will need to have 2.5 mL of the 1% HCl solution and 7.5 mL of distilled water. In mixing the two liquids, you should remember that the order of mixing would be acid to water. So, you use a 10 mL volumetric flask . Put small amount of distilled water and add the 2.5 mL of HCl solution. Lastly, dilute with distilled water up to the 10 mL mark.