Answer:
The true statements are:
The solution is acidic
The pH of the solution is 14.00 - 10.53.
![10^{-10.53}=[OH^-]](https://tex.z-dn.net/?f=10%5E%7B-10.53%7D%3D%5BOH%5E-%5D)
Explanation:
The pH of the solution is defined as negative logarithm of hydrogen ion concentration present in the solution .
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
- The pH value more 7 means that hydrogen ion concentration is less ,alkaline will be the solution.
- The pH value less 7 means that hydrogen ion concentration is more ,acidic will be the solution.
- The pH value equal to 7 indicates that the solution is neutral.
The pOH of the solution is defined as negative logarithm of hydroxide ion concentration present in the solution .
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
The pOH of the solution = 10.53
![10.53=-\log[OH^-]](https://tex.z-dn.net/?f=10.53%3D-%5Clog%5BOH%5E-%5D)
The pH of the solution = ?
Here, the pH of the solution is less than 7 which means that solution acidic.
Answer:
Explanation:
The main task here is that there are some missing gaps in the above question that needs to be filled with the appropriate answers. So, we are just going to do rewrite the answer below as we indicate the missing gaps by underlining them and making them in bold format.
SO; In the quantum-mechanical model of the hydrogen atom.
As the n level increases. the energy <u>increases</u> and thus levels are <u>closer to </u>each other. Therefore, the transition <u>3p→2s</u> would have a greater energy difference than the transition from <u>4p→3p.</u>
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols
For every 1 mol of gas, there will be
24000 cm^3 of gas
Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:
We are given:
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 206.7 yrs
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = ?
[A] = amount left after decay process = 1.40 g
Putting values in above equation, we get:
![0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}](https://tex.z-dn.net/?f=0.0101yr%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B206.7yrs%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B1.40%7D)
![[A_o]=11.3g](https://tex.z-dn.net/?f=%5BA_o%5D%3D11.3g)
Hence, the initial amount of Uranium-232 present is 11.3 grams.
