Br
First of all, As is a metalloid, not a nonmetal. Second of all, as the rows get higher on the periodic table, the elements are more reactive. As you move right on the periodic table, the elements become more reactive.

3 0

3 years ago

Using the periodic table to locate each element, write the electron configuration of(c) Re.

Aleksandr [31]

Rhenium is a chemical element with the symbol Re and atomic number 75. The electron configuration of Re is [Xe] $6s^{2} 4f^{14} 5d^{5}$ .

<h3>How to write an electronic configuration?</h3>

1. Identify the given element and its atomic number from the periodic table.

2. Write the electron configuration by the energy level and the type of orbital first, then the number of electrons present in the orbital as superscript.

The easiest way to write the electronic configuration for any element is by using a diagonal rule for electron filling order in the different subshells according to the Aufbau principle.

The 3 rules for writing the electron configuration in the orbital box diagram are – the Aufbau rule, the Pauli-exclusion rule, and Hund's Rule.

To learn more about electronic configuration, refer

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5 0

2 years ago

The radioactive atom 61/27 co emits a beta particle. write an equation showing the decay

pickupchik [31]

Answer:

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

Explanation:

The unbalanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

60 = 0 + A, so A = 60 - 0 = 60, and

27 = -1 + Z, so Z = 27 + 1 = 28

Your nuclear equation becomes

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{X}$

Element 28 is nickel, so the balanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

7 0

3 years ago