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
Answer:
Rate = 0.001615 Ms-1
Explanation:
2 NO2 + F2 --> 2 NO2F
The reaction is first order with respect to NO2 and also first order with respect to F2.
The rate law is given as;
Rate = k [NO2] [ F2]
k = 1.58E-4 M-1s-1
[NO2] = 2.84 M
[F2] = 3.60 M
Rate = ?
Inserting the values into the equation, we have;
Rate = 1.58E-4 * 2.84 * 3.60
Rate = 0.001615 Ms-1
Answer:
Boiling- 212° F melting- 32°F
Explanation:
