Answer:
3.125% will remain after 125 days
Explanation:
Given data:
Half life of Th-234 = 25 days
Percent of thorium remain after 125 days = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 125 days / 25 days
Number of half lives = 5
At time zero =100%
At 1st half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At third half life = 25%/ 2 = 12.5%
At 4th half life = 12.5% /2 = 6.25%
At 5th half life = 6.25% /2 = 3.125%
Number of moles= mass/ molar mass
Or n=m/MM
n = number of moles
m = mass
MM = molar mass
1) n CuO = 2.4g / 79.54g/mol = 0.03 mol CuO
2) n Cu(NO3)2.xH2O = 7.26 g / 205.6 = 0.035 moles of Cu(NO3)2.xH2O
3) 205.6 g
Cu = 63.5 g
N = 14g
O =16g
H= 1 g
63.5+ (14+(16*3))*2+1*2+16 =205.6 g
4) yes is 188g
5) I don’t know, I assume was 1
Answer: the normal boiling point is the temperature at which the vapour pressure is equal to the standard sea-level atmospheric pressure
Explanation:
the correct answer is option d.
6.67*10-3 hrs
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
