26. B
27. D
28. C
Hope this helped ☺️
Hello!
The mass number in isotope notation is denoted A, the atomic number is denoted as Z, and the element is denoted as X.
In the given isotope, the mass of the isotope is 212 amu, and the atomic number is 82.
We know that the number of electrons, and protons are equal to the atomic number. Therefore, there are 82 protons. Also, to find the number of neutrons, we subtract the atomic number from the atomic mass.
212 - 82 = 130 neutrons
<u>Final answers</u>:
- Atomic Number: 82
- Mass number: 212
- Number of Protons: 82
- Number of Neutrons: 130
Answer:
The equivalent weight of M is approximately 31.8 g
The equivalent weight of N is approximately 27.98 g
Explanation:
The given parameters are;
The percentage of the the metal M in in the chloride = 47.25%
Where by the chemical formula for the metal chloride is MClₓ, we have;
47.25% of the mass of MClₓ = Mass of M = W
Therefore, we have;
0.4725 × (W + 35.5·x) = W
0.4725·W + 0.4725×35.5×x = W
W - 0.4725·W = 16.77·x
0.5275·W = 16.77·x
W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M
The equivalent weight of M = 31.799 ≈ 31.8 g
Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g
The equivalent weight of N = 27.98 g.
Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.
Answer:
A. there is an isotope of lanthanum with an atomic mass of 138.9
Explanation:
By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.
Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.
Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.
Thus, the correct option is:
<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>
