Less stable should be the answer.as I think
I hope this helps you
Answer:
dipole-dipole forces, ion-dipole forces, higher molar mass, hydrogen bonding, stronger intermolecular forces
Explanation:
<em>1. H₂S and H₂Se exhibit the following intermolecular forces: </em><em>dipole-dipole forces </em><em>and </em><em>ion-dipole forces</em><em>.</em> These molecules have a bent geometry, thus, a dipolar moment which makes them dipoles. When they are in the aqueous form they are weak electrolytes whose ions interact with the water dipoles
<em>2. Therefore, when comparing H₂S and H₂Se the one with a </em><em>higher molar mass</em><em> has a higher boiling point.</em> In this case, H₂Se has a higher boiling point than H₂S due to its higher molar mass.
<em>3. The strongest intermolecular force exhibited by H₂O is </em><em>hydrogen bonding</em><em>. </em>This is a specially strong dipole-dipole interaction in which the positive density charge on the hydrogens is attracted to the negative density charge on the oxygen.
<em>4. Therefore, when comparing H₂Se and H₂O the one with </em><em>stronger intermolecular forces</em><em> has a higher boiling point. </em>That's why the boiling point of H₂O is much higher than the boiling point of H₂Se.
Answer:
The enthalpy change of the reaction is -66.88 kJ/mol.
Explanation:
Mass of the solution = m = 100 g
Heat capacity of the solution = c = 4.18 J/g°C
Initial temperature of the solutions before mixing = 
Final temperature of the solution after mixing = 
Heat gained by the solution due to heat released by reaction between HCl and silver nitrate = Q
Heat released due to reaction = Q' =-Q = -334.4 J
Moles of silver nitrate = n
Molarity of silver nitrate solution = 0.100 M
Volume of the silver nitrate solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)
Enthalpy change of the reaction = 
1 J = 0.001 kJ
The enthalpy change of the reaction is -66.88 kJ/mol.
Answer:
a) [H3O+] = 1.00 E-10 M ⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M ⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M ⇒ [OH-] = 1.0 E-9 M
Explanation:
- 14 = pH + pOH
- pH = - Log [H3O+]
a) [H3O+] = 1.00 E-10 M
⇒ pH = - Log(1.00 E-10) = 10
⇒ pOH = 14 - 10 = 4
⇒ 4 = - Log[OH-]
⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M
⇒ pH = 4
⇒ pOH = 10
⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M
⇒ pH = 5
⇒ pOH = 9
⇒ [OH-] = 1.0 E-9 M
Hundreds of them....hope it helps
