Respuesta :
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
Final concentration of C is 3.06 M.
The concentration at the end of the interval is 4.02 M.
Hypothetical reaction:
[tex]4A + 3B ------> C + 2D[/tex]
Given:
The rate of change of the reactant A= -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be 0.08 * 3 = 0.24 M
From the chemical reaction,4 moles of A gives 1 mole of C
0.24 M of reacted A will form [tex]\frac{(0.24 * 1)}{4}[/tex] M of C
Amount of C formed at the end of the 3s interval = 0.06 M. If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
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