Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
35g Mg x 1mol / 24g = 840 mol
Answer : The molecular weight of a substance is 157.3 g/mol
Explanation :
As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.
Mass of solute = 7 g
Mass of solution = 100 g
Mass of solvent = 100 - 7 = 93 g
Formula used :
where,
= change in freezing point
= temperature of pure water = 
= temperature of solution = 
= freezing point constant of water = 
m = molality
Now put all the given values in this formula, we get
Therefore, the molecular weight of a substance is 157.3 g/mol
Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
<em>In which pair do both compounds exhibit predominantly ionic bonding? </em>
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.
There are new ideas and questions that people are trying out and solving. Based off of that, the theories and ideas that we have now will change and evolve with the theories that have been tested and the questions that have been answered.
