Answer:
If the electronegativity difference is greater than 2.0, then the bond is ionic.
Explanation:
For ex:
Na has the electronegativity of 0.93
Cl has the electronegativity of 3.16
we subtract 3.16-0.93
We have 2.23 electronegativity difference between Na and Cl
because 2.23 > 2.0
therefore NaCl compound has an ionic bond.
Answer:
Q = -3980.9 j
Explanation:
Given data:
Mass of sample = 30 g
Initial temperature = 56.7 °C
Final temperature = 25 °C
Specific heat of water = 4.186 j/g.°C
Amount of heat released = ?
Formula:
Q = m.c.ΔT
Q = heat released
m = mass of sample
c = specific heat of given sample
ΔT = change in temperature
Solution:
ΔT = T2 -T1
ΔT = 25 °C - 56.7 °C = - 31.7°C
Q = m.c.ΔT
Q = 30 g × 4.186 j/g.°C × - 31.7°C
Q = -3980.9 j
You start by finding the mol of each
59.9g C x (mol C / 12.01 g C) = 4.98 mol C
8.06g H x (mol H / 1.00 g H) = 8.06 mol H
32.0 g O x (mol O / 16.0 g O) = 2 mol O
So when you set it up you have
C4.98 H8 O2
You divide each by the smallest mol. The smallest mol is 2
C2 H4 O2.5
However you can’t have half a mol in the empirical formula. If the value ends in 0.5, you multiply everything by 2
You’re left with
C2H8O5
The EMPIRICAL formula for lucite is C2H8O5
Note empirical is not the same as chemical formula.
The ionic radius increases.
Each time you move down to the next Period, you are adding an extra shell of electrons.
Each shell is further from the nucleus than the one before it, so the .
The image below illustrates the increase in ionic radius as you go down a Group.
The name is sodium chloride. This is an ionic compound so you use the ionic naming system which is the name of the cation followed by the name of the anion. I hope this helps. Let me know if anything is unclear.
