(A)Nuclear change..............
Answer: kg= 0.37
Explanation:
Use the molality formula.
M= m/kg
Answer:
The pH of the solution will be 7.53.
Explanation:
Dissociation constant of KClO=
Concentration of acid in 1 l= 0.30 M
Then in 200 ml = 
The concentration of acid, HClO=[acid]= 0.006 M
Concentration of salt in 1 L = 0.20 M
Then in 300 ml = 
The concentration of acid, KClO=[salt]= 0.006 M
The pH of the solution will be given by formula :
![pH=pK_{a}^o+\log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%5Eo%2B%5Clog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![pH=-\log[2.8\times 10^{-8}]+\frac{[0.06 M]}{[0.06 M]}](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B2.8%5Ctimes%2010%5E%7B-8%7D%5D%2B%5Cfrac%7B%5B0.06%20M%5D%7D%7B%5B0.06%20M%5D%7D)
The pH of the solution will be 7.53.
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:

Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:

Convert this value to mL:

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
The answer is A. Solids only