The correct answer would be 0.1505 M/s. Given the rate of disapperance of HBr, we can easily calculate the rate of apearance of the products by looking at the coefficients of the substances in the reaction. In this case, for every two moles of HBr, 1 mole of Br2 is being produced so the reaction rate would be:
- 1/2 (r(HBr)) = rBr2
- 1/2 (-0.301) = rBr2
Rate of appearance of Br2 = 0.1505 M/s
i only know that E is one of them
Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
As the given reaction is as follows.
(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.
Thus, formation of 
will increase.
- (b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.
Hence, formation of will decrease with decrease in volume.
- When we increase the mount of
then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.
Thus, we can conclude that formation of will increase then.
<span>So when the chemist combines Ethane (CH3CH3) and Chlorine (Cl2) with the intention of producing Chloroethane (CH3CH2Cl), the other product that's formed in this reaction is 1,2-dichloroethane (ClCH2CH2Cl) also called as Ethylene dichloride with molecular weight of 98.954 g/mol. This is a colorless oily flammable substance that weighs heaver when vaporized.</span>
Answer:
Potassium is more reactive than aluminium, so no reaction takes place. But aluminium is more reactive than copper, so it replaces the copper in copper nitrate
<h3>Explanation:</h3>
More reactive metal compound + less reactive metal
-> no reaction
However
Less reactive metal compound + more reactive metal
-> more reactive metal compound + less reactive metal
This is called substitution reaction where the more reactive metal replaces the less reactive metal in the compound.
