Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3
2 answers:
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Answer:
12 atm
Explanation:
First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K
For this question we must employ the Combined Gas Law: , where
is the initial pressure and
is the new pressure.
We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.
Thus, the filled out Combined Gas Law would be:
=
, where the volume, moles of gas, and R are cancelled out.
We can manipulate this equation to derive the new pressure. We find that
9atm≈0.74885.
This means that
≈9/0.74885≈12 atm
The change in temperature (ΔT) : 56.14 ° C
<h3>Further explanation</h3>Given
Cereal energy = 235,000 J
mass of water = 1000 g
Required
the change in temperature (ΔT)
Solution
Heat can be formulated :
Q = m . c . ΔT
c = specific heat for water = 4.186 J / gram ° C
235000 = 1000 . 4.186 . ΔT