By stirring and increasing temperature, there is an increase in dissolving capacity of the solid solute.
<u>Explanation:</u>
If a solute is added to the solution, it doesn't get dissolve easily then we have to increase the temperature, which in turn increases the movement of the solvent (may be water) and the solute particles, thus increases the dissolving power of the solid solute. One more way is by constant stirring, that is by making more contact among the solvent as well as the solute particles there by increasing the solubility of solid solute.
Answer:
d) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity increases because the arsenic atom loses the electron.
Explanation:
This is an example of a n-type semiconductor. The additional electron introduced to the 'grid' of silicon atoms causes an increase in the conductivity of the silicon. This additional electron is introduced as arsenic loses its extra electron.
Answer:
Explanation:
Molal freezing point depression constant of butanol Kf = 8.37⁰C /m
ΔTf = Kf x m , m is no of moles of solute per kg of solvent .
mol weight of butanol = 70 g
235.1 g of butanol = 235.1 / 70 = 3.3585 moles
3.3585 moles of butanol dissolved in 4.14 kg of water .
ΔTf = 8.37 x 3.3585 / 4.14
= 6.79⁰C
Depression in freezing point = 6.79
freezing point of solution = - 6.79⁰C .
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
