Answer:
6 days
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Half life (t½) =?
Next, we shall determine the decay constant. This can be obtained as follow:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Decay constant (K) =?
Log (N₀/N) = kt / 2.303
Log (100/6.25) = k × 24 / 2.303
Log 16 = k × 24 / 2.303
1.2041 = k × 24 / 2.303
Cross multiply
k × 24 = 1.2041 × 2.303
Divide both side by 24
K = (1.2041 × 2.303) / 24
K = 0.1155 /day
Finally, we shall determine the half-life of the isotope as follow:
Decay constant (K) = 0.1155 /day
Half life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 0.1155
t½ = 6 days
Therefore, the half-life of the isotope is 6 days
Answer:
The average kinetic energy of the particles in the mug decreases. The average kinetic energy of the particles in the coffee increases. Thermal energy from the mug is transferred to the coffee. Thermal energy from the coffee is transferred to the mug.
Answer:
ΔG <0 , ΔH > 0 , ΔS > 0 .
Explanation:
From the data given in question , the reaction is a spontaneous process , hence , the value of change in gibbs free energy would be negative , ΔG <0
And , on dissolution process , the temperature of the water decreases , i.e. , it is an endothermic process , i.e. , the change in enthalphy value is positive , ΔH > 0
And , during the process of dissolution , the ammonia salt break does to ions , i.e. , the randomness increases , hence the ΔS > 0
Ca^2+ and I^-
Na+ and Co3^2-
Ga^3+ and ClO3
Cu^2+ and F-
NH4^- and PO4^3-
Fe2+ and (SO4)^2-
Mg2+ and NO3^-
NH4^+ and NO2^-
K^+ and (C2H3O2)^- {C2H3O2 is acetate}
Na^+ and Cr2O7^2-
