Answer:
V = 12.93 L
Explanation:
Given data:
Number of moles = 0.785 mol
Pressure of balloon = 1.5 atm
Temperature = 301 K
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values.
V = nRT/P
V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm
V = 19.4 L /1.5
V = 12.93 L
Answer:
Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution
Explanation:
As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.
When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.
This gas turn red litmus paper blue.
This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).
If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.
<u>Answer:</u> The molar solubility of 
is 
<u>Explanation:</u>
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The balanced equilibrium reaction for the ionization of calcium fluoride follows:
s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
We are given:
Putting values in above equation, we get:
Hence, the molar solubility of is
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
Density is defined as mass/volume (or m/v).
So,
(126.0 g)/(12.5 cm^3)= 10.08 g/cm^3
If your teacher requires correct significant figures, the answer is 10.1 g/cm^3.
If not, the first answer is fine.
