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!
Answer:
The answer you are looking for is A
Hey there!
Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance PO₄.
1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance H.
6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Balance Ca.
1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Our final balanced equation:
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Hope this helps!
Answer:
fruit cake is the right ans
Wait, do you mind clarifying what you would like answered for the question?
