Answer: B. The molecules move faster and become less dense
Explanation:
Liquid is the state of matter in which particles are less tightly bound as compared to solids as they have weak inter molecular forces as compared to solids.
Thus when a solid is heated, the molecules start moving in random motion due to an increase in the kinetic energy. The molecules move apart as the forces will be weak and thus less molecules will be present per unit volume. Thus it will become less dense.
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➷ 4 x 2 = 8
4 x 1 = 4
4 x 4 = 16
16 + 8 + 4 = 28
The answer is option D. 28
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Answer:
Equilibrium constant expression for
:
.
Where
,
, and
denote the activities of the three species, and
,
, and
denote the concentrations of the three species.
Explanation:
<h3>Equilibrium Constant Expression</h3>
The equilibrium constant expression of a (reversible) reaction takes the form a fraction.
Multiply the activity of each product of this reaction to get the numerator.
is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be
.
Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "
" on the product side of this reaction.
is equivalent to
. The species
appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:
.
That's where the exponent "
" in this equilibrium constant expression came from.
Combine these two parts to obtain the equilibrium constant expression:
.
<h3 /><h3>Equilibrium Constant of Concentration</h3>
In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "
" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:
.
mol = conc × v
= 1.5 × 0.09
= 0.135 moles of HCl
HCl + NaOH > NaCl + H2O
1 mole HCl = 1 mole NaOH
0.135 mol HCl = x
x = 0.135 mol NaOH
mass = mol × molar mass
= 0.135 × 40
= 5.4 g
NaOH = 23 + 16 + 1 = 40 g/mol
I'm not a 100% sure if it's correct