Answer:
1 and 3
Explanation:
The vertical columns (groups) of the periodic table are arranged such that all its elements have the same number of valence electrons. All elements within a certain group thus share similar properties.
Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
- For a weak acid like HF, the dissociation of HF will be:
<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>
[H₃O⁺] = [F⁻].
<em>∵ [H₃O⁺] = √Ka.C,</em>
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>
<em></em>
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>
Answer:
4.7 kJ/kmol-K
Explanation:
Using the Debye model the specific heat capacity in kJ/kmol-K
c = 12π⁴Nk(T/θ)³/5
where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K
Substituting these values into c we have
c = 12π⁴Nk(T/θ)³/5
= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5
= 9710.83(298 K/2219 K)³/5
= 1942.17(0.1343)³
= 4.704 J/mol-K
= 4.704 × 10⁻³ kJ/10⁻³ kmol-K
= 4.704 kJ/kmol-K
≅ 4.7 kJ/kmol-K
So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K
V1 = 445ml V2 = 499ml
T1 = 274 K T2 = ?
By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K
<h3>
Answer:</h3>
3.3 × 10²³ molecules Cu(NO₃)₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- 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>
0.55 mol Cu(NO₃)₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

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

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂