<h3>
Answer:</h3>
28 mol CaF
<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>
[Given] 1.7 × 10²⁵ molecules CaF
[Solve] moles CaF
<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>
28.2298 mol CaF ≈ 28 mol CaF
Answer:
Hydrogen is the smallest chemical element. It has one proton and one electron, making it neutral. However, the ions of hydrogen atom are charged species. Thus, the key difference between hydrogen atom and hydrogen ion is that the hydrogen atom is neutral whereas hydrogen ion carries a charge.
2H₂(g) + O₂(g) ⇄ 2H₂O(l)
Δngas = 0 - (2 +1)
= -3
<h3>
What is Δngas?</h3>
The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.
Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.
<h3>
Definition of equilibrium</h3>
When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.
Learn more about Equilibrium
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North Dakota and Michigan because it has to be 32 Farenhieht or less to snow
Answer:
Average atomic mass = 51.9963 amu
Explanation:
Given data:
Abundance of Cr⁵⁰ with atomic mass= 4.34%
, 49.9460 amu
Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu
Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu
Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu
Average atomic mass = 51.9963 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100
Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100
Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100
Average atomic mass = 5199.632 / 100
Average atomic mass = 51.9963 amu
