Answer:
The value of n in the hydrate formula is 5 ,
.
Explanation:

Mass of hydrate of copper sulfate = 14.220 g
Moles of hydrate of copper sulfate =
Mass of copper sulfate after heating = 8.9935 g
Moles of copper sulfate = 

Solving for n, we get:
n = 5
The value of n in the hydrate formula is 5 ,
.
Magnetism is inversely proportional to distance
this means that if distance increases magnetic attraction force decreases
You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:
PH = Pka + log[acid/base]
Step 1 (Figure out the concentrations):
0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3
Step 2 (Plug into equation):
PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]
PH = 3.69
To find - Identify what kind of ligand (weak or strong), what kind
of wavelength (long or short), what kind of spin (high spin or
low spin) and whether it is paramagnetic or diamagnetic for
the following complexes.
1. [Mn(CN)6]4-
2. [Fe(OH)(H2O)5]2
3. [CrCl4Br2]3-
Step - by - Step Explanation -
1.
[Mn(CN)⁶]⁴⁻ :
Ligand - Strong
Wavelength - Short
Spin - Low spin
Number of unpaired electrons = 1 ∴ paramagnetic.
2.
[Fe(OH)(H₂O)₅]²⁺ :
Ligand - Weak ( both OH⁻ and H₂O )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 5 ∴ paramagnetic.
3.
[CrCl₄Br₂]³⁻ :
Ligand - Weak ( both Br⁻ and Cl⁻ )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 3 ∴ paramagnetic.
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)