Answer:
7,94 minutes
Explanation:
If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>
For the zero-order reactions, concentration-time equation can be written as follows:
[A] = - Kt + [Ao]
where:
- [A]: concentration of the reactant A at the <em>t </em>time,
- [A]o: initial concentration of the reactant A,
- K: rate constant,
- t: elapsed time of the reaction
<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>
Data:
K = 4.2 ×10−3atm/s,
[A]o=[HBr]o= 2 atm,
[A]=[HBr]=0 atm (all HBr(g) is gone)
<em>We clear the incognita :</em>
[A] = - Kt + [Ao]............. Kt = [Ao] - [A]
t = ([Ao] - [A])/K
<em>We replace the numerical values:</em>
t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes
So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).
Explanation:
The water is a polar solvent which means it strongly interact with the ionic species dissolved in it. The reason is due to the formation of Hydrogen bonding. Hence, the ionic species are known to be stabilized by solvation.
The H- bonding is the interaction of H-atom of solvent molecules with the electronegative atom of same molecule or, solute species. Ionic solutes dissolve by forming H- bonding.
But in case n- pentane which is a non polar and lack of H-atom attached to electronegative atom. Hence, there is no H- bonding.
Thus, the energy required to separate Na+ and Cl- ion would be huge in water compared to n-pentane.
Answer:
C. 7.50g
Explanation:
The percent (%) by mass of a solute in a solution refers to the number of grams contained in 100g of solution by that solute. In this case, 5% by mass of pottasium chloride (KCl) means 5g of KCl is contained in 100g of solution.
Therefore, in 150g of solution, there would be:
5g/100g × 150g
= 0.05 × 150
= 7.50g of KCl solute.
Hence, 7.50g of pottasium chloride would be expected to be collected by evaporating 150.0 g of the solution.
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