The correct answer to
the question that is stated above is letter c, <span> the outer electron shell.</span>
Valence electrons occur<span> in the outermost shells of an </span>atom.
>> <span>Valence electrons are </span>electrons<span> that are associated with an </span>atom<span>, and that can participate in the formation of a </span>chemical bond.
Answer:
V₂ = 2.96 L
Explanation:
Given data:
Initial volume = 2.00 L
Initial temperature = 250°C
Final volume = ?
Final temperature = 500°C
Solution:
First of all we will convert the temperature into kelvin.
250+273 = 523 k
500+273= 773 k
According to Charles's law,
V∝ T
V = KT
V₁/T₁ = V₂/T₂
V₂ = T₂V₁/T₁
V₂ = 2 L × 773 K / 523 k
V₂ = 1546 L.K / 523 k
V₂ = 2.96 L
<h3>
Answer:</h3>
P₂ = 0.67 atm
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
Answer is: sodium (Na) and iodine (I₂).
<span>
First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
Answer:
Lead is added to warm dilute nitric acid. When the carbonate has reacted with the warm acid, more carbonate is added until the carbonate is in excess.
Explanation:
