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
6.022*10^23 is the answer
Answer:
Our world will never be the same. Our beaches will be submerged, many towns, cities, states, and even countries will be partially if not completely submerged. All these human cities being flooded will kill thousands of people, pollute the ocean, and lose billions of dollars.
Explanation:
Answer:
The answer to your question is 41.6 g of AgCl
Explanation:
Data
mass of NH₄Cl = 15.5 g
mass of AgNO₃ = excess
mass of AgCl = 35.5 g
theoretical yield = ?
Process
1.- Write the balanced chemical reaction.
NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃
2.- Calculate the molar mass of NH₄Cl and AgCl
NH₄Cl = 14 + 4 + 35.5 = 53.5 g
AgCl = 108 + 35.5 = 143.5 g
3.- Calculate the theoretical yield
53.5 g of NH₄Cl -------------------- 143.5 g of AgCl
15.5 g of NH₄Cl ------------------- x
x = (15.5 x 143.5) / 53.5
x = 2224.25 / 53.5
x = 41.6 g of AgCl