Explanation:
The given reaction is as follows.
Hence, number of moles of NaOH are as follows.
n = 
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 
= 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 
[NaA] =
![[A^{-}] = [NaA] = \frac{0.0025 mol}{V}](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20%5BNaA%5D%20%3D%20%5Cfrac%7B0.0025%20mol%7D%7BV%7D)
Now, we will calculate the 
value as follows.
pH = 
![pK_{a} = pH - log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%3D%20pH%20-%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= 
= 3.42
Thus, we can conclude that of the weak acid is 3.42.
Answer:
Subbituminous coal can form at temperatures as low as 35 to 80 °C (95 to 176 °F) while anthracite requires a temperature of at least 180 to 245 °C (356 to 473 °F).
Sub-types: Cannel coal
Child material class: Lignite
Explanation:
Answer:
<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
Explanation:
- Adding solute to water causes the depression of the freezing point.
<em>ΔTf = Kf.m,</em>
Where,
ΔTf is the change in the freezing point.
Kf is the freezing point depression constant (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>Molality is the no. of moles of solute per kg of the solution.</em>
- <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>
<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>
∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.
<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
I believe it will be A. Density, temperature, and viscosity, can be found in water through tests. Hope this helps! Mark as brainly please!
More than one million Earths could fit inside the Sun if it were hollow. The Sun has a radius of 696.340 km / 432.685 mi and a diameter of 1.39 million km / 864.000 mi.
