Answer:
Element A = Oxygen
Element H =
Element B = Aluminum
Element J = Magnesium
Element C = Selenium
Element L = Carbon
Element D = Sodium
Element Q = Francium
Element F = Antimony
Element R = Calcium
Element G = Chlorine
Element S = Tellurium
Explanation:
Element A is Oxygen because: oxygen 6 valence electrons
; is a gas at room temperature
; and is transported in blood to cells.
Element H is Neon because: Neon is a noble gas
; qppears as red light when charged with electricity (Neon light signs) and it has the second highest Ionization energy of the elements
Element B is Aluminum because: Aluminum is a metal and its ion has charge of +3. It is also located on the borders of the Metalloid staircase
.
Element J is Magnesium because its ion has charge of 2+ and is isoelectronic with Neon because it loses two electrons to now have 10 electrons.
Element C is Selenium because its ion that has a charge of -2 is formed by gaining two electrons in order to have 36 electrons which is isoelectronic with Kr
ypton
Element L is Carbon because carbon has the smallest atomic radius of any member in the Carbon family because it is the first member of the family and atomic radius increases on going down the group.
Element D is Sodium because its ion has charge of +1 and it has 2 inner core levels
, the 1 and 2 energy levels.
Element Q is Francium because it has the largest radius and lowest ionization energy of any element
Element F is Antimony. It is a member of Nitrogen family and has the second highest ionization energy level in family
.
Element R is calcium because its on has charge of +2 which is isoelectronic with Argon
. Calcium also has atomic radius is larger than Ar
gon.
Element G is Chlorine. It has the second to the smallest radius of elements in the 3rd period as the second to the last element in the period because atomic radius decreases across a period from left to right.
Element S is Tellurium. It has atomic mass larger than Iodine just to the right of it and is found in the 5th period
<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
Answer:
<em>Sodium dihydrogen phosphate + calcium carbonate</em>
<u>Full ionic equation</u>
2 Na⁺(aq) + 2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)
<u>Net ionic equation</u>
2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)
<em>Sodium oxalate + calcium carbonate</em>
<u>Full ionic equation</u>
2 Na⁺(aq) + C₂O₄²⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + CaC₂O₄(s)
<u>Net ionic equation</u>
C₂O₄²⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + CaC₂O₄(s)
<em>Sodium hydrogen phosphate + calcium carbonate</em>
<u>Full ionic equation</u>
2 Na⁺(aq) + HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + 2 Na⁺(aq) + CO₃²⁻(aq)
<u>Net ionic equation</u>
HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + CO₃²⁻(aq)
Explanation:
Let's consider two kind of equations:
- Full ionic equation: includes all ions and species that do not dissociate in water.
- Net ionic equation: includes only ions that participate in the reaction (<em>not spectator ions</em>) and species that do not dissociate in water.
