I believe the charge is positive as losing an electron suggests that the atom lost a negative charge, leaving it with more positive than negative.
I haven't taken this in a few years very sorry if i'm wrong tho.
Answer:
1) Has a smaller radius than oxygen because of the increased electromagnetic attraction of the nuclei
2) Has a smaller radius than chlorine because all the electrons of F have lower energy levels and have less repulsion of other electrons and hence are more attracted to the nuclei .
Explanation:
Further the electrons are from the nuclei , the bigger the atomic radius is.
(+) attraction of electrons to the nuclei, (-) repulsion of the electrons away from the nuclei.
1) From O to F:
(+) there is one more proton --> Stronger positive charge of the nuclei means that the electrons are attracted more , then they come closer to it and therefore the radius decreases
(-) There is one more electron --> Every electron is repulsed by others away from the nuclei --> the radius increases. But this effect is not so strong because the new electron is added at the same energy level.
Overall the (+) effect is stronger than the (-) effect --> Radius decreases from O to F
2) From F to Cl
(+) there is one more protons --> Same effect as before
(-) There is one more electron --> Every electron is repulsed by others away from the nuclei. But this time the new electrons have a higher energy level --> Meaning that they are less attracted and hence the radius increases.
And also the other inner layers of electrons (electrons of lower energy levels) repulse this new external layer of electrons more effectively than the case of O --> Strong repulsion effect (called electron shielding effect) --> Radius increases
Overall the (-) effect is stronger than the (+) effect --> Radius increases from F to Cl (or decreases from Cl to F)
The bond<span> length in a </span>molecule<span> of </span>KBr<span> is measured to be 282.1 pm, compared with a </span>nonpolar<span> covalent radius sum for K and Br of 310,4. </span>
Answer:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M
<span>under specific conditions it can promptly be taken apart easily, yielding a great deal of energy . It is subsequently said to be 'fissile' and we utilize the articulation 'atomic splitting'. In the interim, similar to every single radioactive isotope, they rot.</span>