Agitation, or stirring of the solute in the solvent increases the solubility of the solution
Answer : (C) Hafnium is the most likely identity of the given substance.
Solution : Given,
Mass of given substance (m) = 46.9 g
Volume of given substance (V) = 3.5 
First, find the Density of given substance.
Formula used :
Now,put all the values in this formula, we get
= 13.4 g/
So, we conclude that the density of given substance (13.4 g/) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ respectively).
According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.
Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.
So, Hafnium is the most likely element which is the identity of the given substance.
Answer:
This isotope has 59 electrons giving it a charge of -2.
Explanation:
To find this we have to understand isotope relates to the mass of the nucleus. This isotope has 59 electrons to counter the protons and give it a negative charge.
Answer:
ΔH rxn = -1010 kJ/molC₂H₂
Explanation:
To obtain the enthalpy change for a reaction from bond energies what we do is to make an inventory of the bonds broken and formed for the balanced chemical reaction:
C₂H₂ + 5/2O₂ ⇒ 2CO₂ + H₂O
Bond Broken Bonds Formed
2 C-H + 1 C≡C + 5/2 O=O 4C=O + 2 H-O
Enthalpy bonds broken:
2 mol (456 kJ/mol)+ 1 mol (962 kJ/mol) + 5/2 mol (499 kJ/mol) = 3121.5 kJ
Enthalpy bond formed:
4 mol (802 kJ/mol) + 2 mol (462 kJ/mol) = 4132.0 kJ
ΔH rxn = H broken - H formed = 3121.5 kJ - 4132.0 kJ = - 1010 kJ (per mol C₂H₂ )
