<h3>
Answer:</h3>
12.4 g N₂O₂
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 1.24 × 10²³ molecules N₂O₂
[Solve] grams N₂O₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of N₂O₂ - 2(14.01) + 2(16.00) = 60.02 g/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>
12.3588 g N₂O₂ ≈ 12.4 g N₂O₂
Answer:
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Explanation:
P 15 atm
V₁ = 12x103 m3 -3 3 V₁ = 12x10 m" T₁ = 25°c T= 28815 K
P₁ = 1.52x10 Pa
Now, Finel quantities, T₂ = 35°C = 308.15 °K
V2= 851 = -3 3
Now
by ideal gas equation
P2Y2 T2
= PV, T2
1152 X106 x 12x16 3 x 308115 8,5×103 x 288,15 = Pa
=
21294×10 Pa
P₂ = 21264 Atm
final
pressure will be
21264 Adm.
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<span>All 4 C-H bonds are identical.
Hybridisation or hybridization is the idea of blending atomic orbitals into new hybrid orbitals, with various energies, shapes, and so on, than the component atomic orbitals, appropriate for the matching of electrons to make chemical bonds in valence bond theory.
</span>