An atom that has 13 protons and 15 neutrons is isotope of Aluminium (answer C)
<u><em>Explanation</em></u>
- Isotope is a form of the same element with the equal number of protons but difference number of neutrons in their nuclei.
- In other words isotope has the same atomic number but different mass number.
- Atomic number of a element is determined by number of protons of an element.
- from the periodic table Aluminum in atomic number 13 therefore it has 13 protons <em>therefore an atom that has 13 protons and 15 neutrons is a isotope of Aluminium. </em>
Answer:
Lewis structure in attachment.
Explanation:
Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an <u>expanded octet</u>.
The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.
I would say D. Let me know if i am wrong.
2n² rule:
This rule is used to determine number of electrons in particular shell.
n=1 means K shell
n=2 means L shell
n=3 means M shell
n=4means N shell
The first K shell can hold upto 2 electrons, L shell can hold up to 8, third M shell can hold up to 18and the fourth N shell can hold upto 32 electrons. This rule of arrangement of electrons according to the shell is known 2n2 rule where n means number of shell.
For example: There is one proton in the nucleus of hydrogen atom, which moves in K shell path. It has no neutron.
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
