The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.
1 answer:
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464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.
Explanation:
From the question data given is :
volume of strontium in sea water= 9.84x10^8 cubic meter
(1 cubic metre = 1000000 ml)
so 9 .84x10^8 cubic meter
= 984 ml.
density of sea water = 1 gram/ml
from the formula mass of strontium can be calculated.
density =
mass = density x volume
mass = 1 x 984
= 984 grams of strontium will be recovered.
98400 centigram of strontium will be recovered.
Strontium is an alkaline earth metal and is highly reactive.
<h3>Answer:</h3>
Temperature is 529.164 K
<h3>Explanation:</h3>We are given
Number of moles of Ne (n) = 0.019135 moles
Volume (V) = 878.3 mL
Pressure (P) = 0.946 atm
We are required to calculate the temperature;
We can do this using the ideal gas law equation which is;
PV = nRT, where P is the pressure, n is the number of moles, V is the volume, R is the ideal gas constant (0.082057 Latm/mol/K) and T is the temperature.
From the equation;
Therefore, the temperature will be 529.164 K.