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.
Answer:
A atom is a small unit to which matter can be divided without releasing electrically charged particles
Explanation:
Answer:
0.18× 10²³ molecules
Explanation:
Given data:
Mass of copper hydroxide = 3.30 g
Number of molecules = ?
Solution:
Number of moles = mass/molar mass
Number of moles = 3.30 g/97.56 g/mol
Number of moles = 0.03 mol
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
0.03 mol × 6.022 × 10²³ molecules / 1mol
0.18× 10²³ molecules
Answer:
The molar mass of the liquid 62.89 g/mol
Explanation:
Step 1: Data given
Mass of the sample = 0.1 grams
Temperature = 70°C
Volume = 750 mL
Pressure = 0.05951 atm
Step 2: Calculate the number of moles
p*V = n*R*T
n = (p*V)/(R*T)
⇒ with n = the number of moles gas = TO BE DETERMINED
⇒ with p = The pressure = 0.05951 atm
⇒ with V = The volume of the flask = 750 mL = 0.750 L
⇒ with R = The gasconstant = 0.08206 L*atm/K*mol
⇒with T = the temperature = 70 °C = 343 Kelvin
n = (0.05951 *0.750)/(0.08206*343)
n = 0.00159 moles
Step 3: Calculate molar mass
Molar mass = mass / moles
Molar mass =0.1 gram / 0.00159 moles
Molar mass = 62.89 g/mol
The molar mass of the liquid 62.89 g/mol
