Answer:
The energy released in the decay process = 18.63 keV
Explanation:
To solve this question, we have to calculate the binding energy of each isotope and then take the difference.
The mass of Tritium = 3.016049 amu.
So,the binding energy of Tritium = 3.016049 *931.494 MeV
= 2809.43155 MeV.
The mass of Helium 3 = 3.016029 amu.
So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV
= 2809.41292 MeV.
The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV
1 MeV = 1000keV.
Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.
So, the energy released in the decay process = 18.63 keV.
B its viscosity decreases
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane() = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
B and D is out. It cant be A because heat of combustion is substance not compound. So the answer is D.
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)
The specific heat of the metal is 0.44 J/ (°C × g)