It would be “To obey the law of conservation of mass”
<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
30% should be the percentage of oxygen if the total mass of fe2o3 is 160.
Moles of potassium permanganate = 0.0008
<h3>Further explanation </h3>
Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range
Reaction
5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)
The end point ⇒titrant and analyte moles equal
titrant : potassium permanganate-KMnO4
analyte : sodium oxalate - Na2C2O4
so moles of KMnO4 = moles of Na2C2O4
moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!