How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at s
1 answer:
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Answer:
Option 4 is correct (72 seconds)
Explanation:
Option 4 is correct (72 seconds)
The formula we are going to use is:
Where:
A is the final concentration
A_o is the initial concentration
k is the constant
t is the time
Half-Life=0.693/k
Half-life in our case=24 seconds
k=0.693/24
k=0.028875 s^-1
Since the concentration is decreased by 87.5 % which means only 12.5%(100-87.5%) is left.
The ratio will become 0.125
t≅ 72 seconds
Answer:
319.15^{o}C[/tex]
Explanation:
When all other variables are constant, we are allowed to use the formula
=
Answer : The unit for the rate constant in the rate law for the formation of ammonia is,
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
Rate law expression for the reaction:
where,
a = order with respect to
b = order with respect to
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Dividing 2 by 1, we get:
Dividing 3 by 1 and also put value of b, we get:
Thus, the rate law becomes:
Now, calculating the value of 'k' by using any expression.
The value of the rate constant 'k' for this reaction is
That means, the unit for the rate constant in the rate law for the formation of ammonia is,