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:
The volume of the gas will be 78.31 L at 1.7 °C.
Explanation:
We can find the temperature of the gas by the ideal gas law equation:
Where:
n: is the number of moles
V: is the volume
T: is the temperature
R: is the gas constant = 0.082 L*atm/(K*mol)
From the initial we can find the number of moles:
Now, we can find the temperature with the final conditions:
The temperature in Celsius is:
Therefore, the volume of the gas will be 78.31 L at 1.7 °C.
I hope it helps you!
Answer: After three half-lives 1/8 (12.5%) of the original sample remains
Answer: b.) they tend to lose electrons to gain stability
Explanation:
