Answer: The element shown in the image is Helium (He).
Explanation: We are given a image of an atom having protons, neutrons and electrons.
Number of protons as shown in image = 2
Number of neutrons as shown in image = 2
Number of electron as shown in image = 2
Atomic number = Number of protons = Number of electrons
Atomic number of the element = 2
Atomic Mass = Number of protons + Number of neutrons
Atomic mass = 2 + 2 = 4
The element having Atomic number = 2 and mass number = 4 is Helium.
Element = 
Answer:
The answer I believe is B. Erosion
Explanation:
Just sounds better than all the other choices.
No, because hydrogen isn’t brought out of the equation
Answer: Metals form cations.
The alkali metals (the IA elements) lose a single electron to form a cation with a 1+ charge.
The alkaline earth metals (IIA elements) lose two electrons to form a 2+ cation.
Aluminum, a member of the IIIA family, loses three electrons to form a 3+ cation.
Therefore, metals in the s and p block of the periodic table have 1, 2 or 3 electrons in their outermost orbit (or valence shell). Now to gain a stable octet metals lose either 1, 2 or 3 electrons from the valence shell thus forming cation with +1, +2 or +3 charge.
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
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