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:
please brainlist answer
Explanation:
The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.
The volume could be calculated by using <span>V = n RT / P </span>
In which V = Volume
n = number of Moles
R= The Gas constant
T = Temperature (ideally this would be in Kelvin, but i don't see it in the option)
P = Pressure
I believe the answer is
<span>V = (1.5mol) (0.08205 L*kPa/K*mol) (22Celsius)/100 kPa
</span>
Answer:
A positive ions is always smaller than the corresponding atom.
A negative ion is always larger than the corresponding atom.
Explanation:
The reason for this is that, when a positive ion is formed, a full shell is usually removed with its electrons thereby reducing the size of the electron cloud and decreasing the size of the electron cloud.
A negative ion is formed by addition of more electrons to the electron cloud hence it spreads out. Interelectronic repulsion accounts for the larger size of the negative ion.
