Answer:
there must be free-moving electrons in a compound to conduct electricity.
in solid ionic compounds, there are no free electrons so it can't conduct electricity.
But in aqueous form the forces of attraction are broken and the ions (cations and anions) are free to move. This means there are free electrons that allow the aqueous compound to conduct electricity.
Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
Answer:
Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)
Explanation:
Elements are placed into different groups in the periodic table. Elements in the same group are those that have the same number of valence electrons in their outermost shell and as a result will behave similar chemically i.e. will react with other elements in the same manner.
Chlorine and Bromine are two elements belonging to group 7 of the periodic table. They are called HALOGENS and they have seven valence electrons in their outermost shell. Hence, when they form a compound with Calcium, a group two element, these compounds (CaCl2 and CaBr2) will possess similar properties because they have elements that are from the same group (halogen group).
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
