When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer
1 answer:
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Answer : The internal energy change is, -506.3 kJ/mol
Explanation :
Formula used :
or,
where,
= change in enthalpy =
= change in internal energy = ?
= change in moles
Change in moles = Number of moles of product side - Number of moles of reactant side
According to the reaction:
Change in moles = 0 - 2 = -2 mole
That means, value of = 0
R = gas constant = 8.314 J/mol.K
T = temperature =
Now put all the given values in the above formula, we get
Therefore, the internal energy change is -506.3 kJ/mol
Answer:
The answer is D. gamma rays
Explanation:
A radioactive atom can have three different types of emission:
alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.
Beta particles (β) = they have 5x amu and they have an intermediate penetrating power
Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher
For this reason Gamma emissions (γ) has the smallest mass value.
Answer: P₂=0.44 atm
Explanation:
For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.
Gay-Lussac's Law:
First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.
64+273K=337K
Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.
Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.
P₂=0.44 atm