The most suitable answer is C becuase they would gain two elctrons to atain that stable OCTET thus becoming a anion with a charge of -2 and by virtue oxidation states of -2. There is however an exception with oxygen in two cases. But I still remain that the best answer would be C
Answer:
The calculated density will be larger
Explanation:
The calculated density will be <u>larger</u>. Because, the volume is taken accurately, by the water displacement method. But, when we the took the mass, the water was present on the unknown solid. So, the mass of that water was added to the original mass of the solid. Hence, the mass measured was larger than the original mass. We, know from the formula of density that density is directly proportional to the mass of the object.
Density = Mass/Volume
Hence, the larger measured mass means the larger value of density.
The less soluble salt : PbCl₂
<h3>Further explanation</h3>
Given
0.1 M NaCl
Required
The less soluble salt
Solution
If we see from the answer option, the salt that is more difficult to dissolve in NaCl is PbCl₂ because it has the same ion (Cl)
When PbCl₂ is dissolved in water, ionization will occur
PbCl₂ ⇒ Pb²⁺+ 2Cl⁻
So, when dissolved in NaCl, NaCl itself will be ionized
NaCl ⇒ Na⁺ + Cl⁻
Based on the principle of equilibrium, the addition of an ion (one of the ions is enlarged), the reaction will shift towards the ion that was not added. In addition to this Cl ion, the reaction will shift to the left so that the solubility of PbCl₂ will decrease (the reaction to the right decreases)
iodine which should appear before tellurium and argon which should appear after potassium
There is a 3rd one: nickel should appear before cobalt
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂
