Answer:
Elements
Explanation:
The <u> elements</u> which are represented by a single uppercase letter, or represented by an uppercase letter followed by a lowercase letter. This tells you the types of <u>elements</u> in the compound.
Elements are one of the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means. They are made up of small indivisible particles called atoms, all having the same number of protons.
Answer:
81 L gas
General Formulas and Concepts:
<u>Ideal Gas Law</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 3.6 mols gas at STP
[Solve] volume (L) of gas
<u>Step 2: Convert</u>
- [DA] Set up:

- [DA] Multiply [Cancel out units]:

<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
80.64 L gas ≈ 81 L gas
The activation energy barrier is 40.1 kJ·mol⁻¹
Use the Arrhenius equation






Answer:

Explanation:
Manganese is the element of group 7 and forth period. The atomic number of Manganese is 25 and the symbol of the element is Mn.
The electronic configuration of the element, manganese is -

To form
, it will lose 3 electrons from the valence electrons and thus the configuration of the ion is:-
