Answer:
17 protons, 20 neutrons, and 17 electrons.
Explanation:
A periodic table can be defined as the standard arrangement of chemical elements by atomic number, electronic configuration and chemical properties in a tabular form.
Generally, a proper representation of the mass number and atomic number of chemical elements is key and very important in chemistry.
Furthermore, as a rule, it should be noted that the mass number (nucleon number) is always larger than the atomic number(number of proton).
The mass number of this neutral atom of Cl-37 is 37 and we know that the atomic number (number of protons) of chlorine is 17. Also, the atomic number of an element is equal to the number of its electrons.
A neutral atom of Cl-37 has 17 protons, 20 neutrons, and 17 electrons.
Hence, a neutral atom of Cl-37 can be identified based on its number of protons because it represent its atomic number, which is what is used to differentiate an atom of an element from the atom of another chemical element.
The answer is: [B]: "ionic salt" .
___________________________________________________
Note: There is no "sharing of electrons" among the elements in this compound; so this compound in NOT a "covalent molecule".
However, there is ionic bonding: Cu²⁺ and Cl⁻ ; to form: "CuCl₂" .
____________________________________________________
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
