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).
The question is incomplete. Complete question is attached below:
...........................................................................................................................
Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.
<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
