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!
You’ll need to be sure to count all the atoms in each side of the chemical equation.
Answer:
The most common example is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6 L and a pressure of 2.1 - 18914… ... of 5.6 L and a pressure of 2.1 atm, what is the initial volume of the gas? ... An ideal gas is at a pressure of 1.4 atm and has a volume of 3 L.
Explanation:
I hope I help :)
Theoretical yield is the quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction. It is the amount of product resulting from a perfect chemical reaction and thus not the same as the amount you'll actually get from a reaction.
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
