Answer:
1 .
2.
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which
behaves as cation and
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation
is greater than that of
and also the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
Answer:
C
Explanation:
Temperature is directly related to kinetic energy (KE). As we raise temperature, we are raising KE, as well. Particles with more KE move more quickly and with more force.
This means that these particles are more likely to collide with each other and react to allow the chemical reaction to follow through. In turn, if the chemical reaction is more likely to go to completion, the reaction rate increases, eliminating A and B.
The concentration of the solute is not affected by the temperature; in other words, temperature will not increase or decrease the amount of solute in the solution, so eliminate D.
Thus the answer is C.
Hope this helps!
Answer:
Assuming it was collected from the atmosphere it would be virtually nothing
Explanation:
hydrogen makes up 0.000055% of the atmosphere while oxygen makes up 23 percent. 20/400000 cm^3 of hydrogen
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
<em>Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.</em>
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
<em>Moles CH3COOH: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
<em>Moles CH3COONa: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
<h3>pH = 4.71</h3>