Answer:
Fewer bubbles will be produced because of fewer collisions of reactant molecules
Explanation:
As the solid dissolves into the solution after the liquid has been vigorously bubbled, if the temperature of the liquid is reduced a little, what will happen is that fewer bubbles will be produced as a result of lesser amount of collisions occurring between the reactant molecules
At STP condition 1 mol of any ideal gas will have a volume of 22.4L
1.75 mol of F2 x 22.4 L / 1 mol = 39.2 L
Answer:
1st column group 1 - group2
second column Halogen group - zero group ( inert gases )
third column - very active metals , tend to lose one electron -less reactive than group one , tend to lose 2 electrons - very active nonmetals, tend to gain one electron - Nobel gases , do not gain or lose electrons
Explanation:
Explanation:
When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral. ... Chemically, we say that the atoms have formed bonds.
Lewis structure for each of the following N₂O₃ with no N¬N bond is attached below.
Even though pi symmetry occupies the antibonding orbitals of NO, this is unimportant after the dimer forms. A sigma connection exists. The enthalpy of the newly formed sigma bond in the dimer is low because the loss of a particularly distinctive set of single-electron resonance forms that were available for no monomer offset the net gain in bond. When the whole free energy is taken into account, there is no gain because the entropic effects are on the order of 1030kJ/mol, and dimerization is entropically disfavored at G=17kJ/mol. Therefore, any little increase in enthalpy is cancelled out by the loss of entropy.
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