The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
Answer:
[H+] = 1.74 x 10⁻⁵
Explanation:
By definition pH = -log [H+]
Therefore, given the pH, all we have to do is solve algebraically for [H+] :
[H+] = antilog ( -pH ) = 10^-4.76 = 1.74 x 10⁻⁵
Answer: A supersaturated solution will not contain undissolved solute because the undissolved solute will be indicative of saturated solution.
Explanation:
A supersaturated solution is the one that consists of more than the maximum concentration of the solute in the solvent that is being dissolved at a given temperature. A saturated solution is the one in which the maximum concentration of solute has been dissolved in the solvent and no additional solute can be dissolved further.
According to the given statement, a solution with undissolved solute is a saturated solution rather a supersaturated solution.
To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.
