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:
Exposure of silver chloride to sunlight for a long duration turns grey due to photolytic decomposition i.e decomposition in the presence of sunlight.
Explanation:
When silver chloride, AgCl is exposed to sunlight for a long time, it will undergo decomposition as the sunlight provides sufficient energy needed to decomposed the salt, AgCl to metallic silver and chlorine gas. This can be seen in the equation below:
2AgCl —> 2Ag + Cl2
Answer:
Those individuals with advantageous traits are more likely to survive and reproduce. The survivors pass down these advantageous traits to their offspring.
Explanation:
Answer:
a) the minimun of acetic anhydride required for the reaction is 2.175 g (CH3CO)2O
b) V acetic anhydride = 2.010 mL
Explanation:
C6H4OHCOOH + (CH3CO)2O ↔ C9H8O4 + C2H4O2
⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH
⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O / mol C6H4OHCOOH ) = 0.0213 mol (CHECO)2O
⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O
b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O
