Answer:
Molarity of NaOH = 1.8 M.
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Volume = 500 mL
Molarity of NaOH =?
Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 36 / 40
Mole of NaOH = 0.9 mole
Next, we shall convert 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Finally, we shall determine the molarity of NaOH. This can be obtained as follow:
Mole of NaOH = 0.9 mole
Volume = 0.5 L
Molarity of NaOH =?
Molarity = mole / Volume
Molarity of NaOH = 0.9 / 0.5
Molarity of NaOH = 1.8 M
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.
<span>The half-life of a first-order reaction is determined as follows:
</span>t½<span>=ln2/k
From the equation, we can calculate the </span><span>first-order rate constant:
</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³
When we know the value of k we can then calculate concentration with the equation:
A₀ = 2 g/100 mL
t = 2.5 h = 150min
A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml
= 6.3 × 10⁻⁴ mg / 100ml
The hydroxide concentration is written as [OH-]. A useful equation is that pH + pOH = 14. Since the pH is 4, the pOH must be 10. To get [OH-], again take the neg. anti-log of both sides: - anti log (pOH-) = - antilog(10) ---> [OH-]=1 x 10^-10 M or 1 x 10^-10 moles [OH-] per liter solution (in this case, lake).
The hydroxide ion concentration describe in question 3 is 10-10 M
