Answer:
the correct answer is Group of answer choices
Answer:
The mass defect of a deuterium nucleus is 0.001848 amu.
Explanation:
The deuterium is:
The mass defect can be calculated by using the following equation:
![\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20%5BZm_%7Bp%7D%20%2B%20%28A%20-%20Z%29m_%7Bn%7D%5D%20-%20m_%7Ba%7D)
Where:
Z: is the number of protons = 1
A: is the mass number = 2
: is the proton's mass = 1.00728 amu
: is the neutron's mass = 1.00867 amu
: is the mass of deuterium = 2.01410178 amu
Then, the mass defect is:
![\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20%5B1.00728%20amu%20%2B%20%282-%201%291.00867%20amu%5D%20-%202.01410178%20amu%20%3D%200.001848%20amu)
Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.
I hope it helps you!
Explanation:
Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.
According to Boyle's law: The pressure of the gas is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).
(At constant temperature)
The pressure beneath the sea is 202 kPa and the atmospheric pressure is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).
Answer:
20.5torr
Explanation:
Given parameters:
V₁ = 15L
P₁ = 8.2 x 10⁴torr
V₂ = 6 x 10⁴L
Unknown:
P₂ = ?
Solution:
To solve this problem we have to apply the claims of Boyle's law.
Boyle's law is given mathematically as;
P₁ V₁ = P₂V₂
where P₁ is the initial pressure
V₁ is the initial volume
P₂ is final pressure
V₂ is final volume
8.2 x 10⁴ x 15 = P₂ x 6 x 10⁴
P₂ = 20.5torr