Answer:
24.9 L Ar
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Aqueous Solutions</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 40.0 g Ar
[Solve] L Ar
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ar - 39.95 g/mol
[STP] 22.4 L = 1 mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
24.9235 L Ar ≈ 24.9 L Ar
The amount of Al that would be needed will be 0.79 grams
<h3>Stoichiometric calculations</h3>
From the equation of the reaction below:

The mole ratio of Al to
is 2:3.
Mole of 71.8 mL. 0.610 M
= 0.610 x 71.8/1000 = 0.0438 moles
Equivalent mole of Al = 2/3 x 0.0438 = 0.029 moles
Mass of o.o29 moles Al = 0.029 x 26.98 = 0.79 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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You proteins and electrons have to be balanced like if you have 18+ you need 18- for it to be stable I hope you understand