The intermolecular forces that are responsible for the dissolution of Ethylene glycol in water is hydrogen bonding dipole-dipole forces and dispersion forces.
Both ethylene glycol and water contains the pair of hydrogen and oxygen.
The hydrogen of one atom create a bond with the oxygen of other atom this results in the formation of intra molecular hydrogen bonding.
The electron are non uniformly distributed over the molecule or the atom which results in the fluctuation of the electron density in the atom.
So it creates are dispersion forces which is present all over the molecule this forces helps to increase the strength of the bond formed between the ethylene glycol and water because they have large masses.
Both ethylene glycol and water are polar molecules because of being polar they form dipole and the dipole of both the molecules interact with each other in order to form bond between the atoms which eventually results in the formation dissolution of ethylene glycol in water.
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Compete Question - which intermolecular forces are responsible for the dissolution of ethylene glycol? select all that apply hydrogen bonding, dipole-dipole, dispersion and Ion dipole interaction.
The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂<span>O(l)
</span>
n=mass in g/M.M
15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂
no. of mol of HCl:
n=0.5 mol/L*0.075L=0.0375 mol
This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.
Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.
Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂
no. of mol of CaCl₂= 0.0375/2= 0.01875 mol
mass in g=n*MM= 0.01875*111= 2.08 g
B I’m pretty sure. (Sorry if I wrong-)
Answer: __3__Na2CO3 + __2__Ag3P → __2__Na3P + __3__Ag2CO3
Explanation:
In balancing equations you need to make sure the number of atoms before the reaction should be equal to the number of atoms after the reaction.
Before the reaction: We had 6Na, 9CO, 6Ag and 2p
After the reaction: We have 6Na, 9CO, 6Ag and 2p
And now we are sure that atoms before the reaction are equal to atoms after the reaction
