Answer : The oxygen will appears in the final chemical equation as a reactant.
Explanation :
The given chemical reactions are:
(1) 
(2) 
Now we are adding reaction 1 and reaction 2, we get the final chemical equation.
The final chemical equation is:
Thus, the oxygen will appears in the final chemical equation as a reactant.
835 g Ar to liters of Argon gas is calculated as follows
find the moles of Ar = mass/molar mass
= 835g /39.95 g/mol = 20.9 moles
At STP 1mole of ideal gas = 22.4 L, what about 20.9 moles
= 20.9 moles/1mole x 22.4 L =468.16 L of Argon
Answer:
0.1440M
Explanation:
Let''s bring out the parameters we were given;
Rate constant = 8.74 x 10^-4s^-1
Initial Concentration [A]o = 0.330M
Final concentration [A]= ?
Time = 800s
The reaction is a first order reaction, due to the unit of the rate constant. In first order reactions, the reaction rate is directly proportional to the reactant concentration and the units of first order rate constants are 1/sec.
Formular relating these parameters is given as;
ln[A] = ln[A]o − kt
Making [A] subject of interest, we have;
ln[A] = ln[A]o − kt
ln[A] = ln(0.330) - ( 8.74 x 10^-4 * 800)
In[A] = - 1.1086 - (6992 x 10^-4)
ln[A] = -1.8079
[A] = 0.1440M
The answer you’re looking for is
C.
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