Answer:
"1.
buoyant or suspended in water or air.
2.not settled in a definite place; fluctuating or variable."
Explanation:
Hope this helps! :)
Here's the equation you use: Density = mass/volume
1) 5.2g/cm^3 = m/3.7cm^3
2) m = 5.2g/cm^3 x 3.7cm^3
3) m = 19.24g
You can check the answer by plugging it in
19.24g/3.7cm^3
= 5.2g/cm^3
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
First the theoretical yield of Nabr
by use of mole ratio between FeBr3 and NaBr which is 2:6 the theoretical yield
=2.36 x6/2= 7.08 moles
the % yield = actual yield/ theoretical yield x 100
that is 6.14/7.08 x100= 86.72%
Answer:
Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
Explanation:
First, we will write the molecular equation because it is the easiest to balance.
FeCl₃(aq) + 3 KOH(aq) → Fe(OH)₃(s) + 3 KCl(aq)
The full ionic equation includes all the ions and the molecular species.
Fe³⁺(aq) + 3 Cl⁻(aq) + 3 K⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s) + 3 K⁺(aq) + 3 Cl⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
