Answer:
6.Given,
Final Velocity =60m/s
Initial Velocity= 0
Time=10 sec
A=?
A=Final Velocity- Initial Velocity/time
=60-0/10
=60/10
=6m/s ans.
Explanation:
Acceleration = Final Velocity - Initial Velocity/Time
By using this Formula we can calculate Acceleration.
The picture of correct answer is attached it is the second one from the left since it has 5 electrons in outermost shell so it can share 3 electrons to complete the octet rule, while first one contains only one electron in outermost shell so it easier to be lost, third one has 8 electrons so it has complete valence shell and last one with only two electrons in outermost shell so it is easier to lose these two electrons.
from 5 to 7 electrons on outermost shell can form covalent bond
<span>(88.39 / (88.39 + 44.61) ) x (5264000) = answer for copper grams
(44.61 / (88.39 + 44.61)) x (5264000) = answer for sulfur grams
Sulfur is a non-metal used to make paper, number 16 on periodic table.</span><span>
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Only C is correct.
A will produce either an methylbromide + alcohol
B will produce alcohol and alkyl bromide as well
C cyclic ether when reacted with HBr will only produce 1 product which has alcohol group (-OH group) on one end and Bromide group on the other end
Hello!
To find electron configuration for Idoine we need to understand the following steps:
- Finding the Atom's Atomic Number (tells us the specific number of electrons)
- Determining the Charge of the Atom
- Understanding the orbitals (Set S [Contains 2], P [Contains 3, Holds 6], D [Contains 5, Holds 10], F [Contains 7, Holds 14], and there are some theoritical ones.) [Overall the sets go SPDFGHIK
- Understanding notations in configuartion. The notations display the number of electrons in the atom and set.
In this case, for Iodine. If we follow these rules we can see that the electron configuration is [Kr] 4d^10 5s^2 5p^5. We use Krytpon in front because that is the last full and stable noble gas before this particular element. Atoms are just trying to be stable so the goal is to achieve that full shell.