I think it's C. Please mark brainliest
Answer:
0.246 kg
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist adds 370.0mL of a 2.25 M iron(III) bromide (FeBr₃) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>
<em />
We have 370.0 mL of 2.25 M iron(III) bromide (FeBr₃) solution. The moles of FeBr₃ are:
0.3700 L × 2.25 mol/L = 0.833 mol
The molar mass of iron(III) bromide is 295.56 g/mol. The mass corresponding to 0.833 moles is:
0.833 mol × 295.56 g/mol = 246 g
1 kilogram is equal to 1000 grams. Then,
246 g × (1 kg/1000 g) = 0.246 kg
<h3>
Answer:</h3>
7.4797 g AlF₃
<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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol KF = 2 mol AlF₃
Molar Mass of K - 39.10 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃
Answer:
One possible reason why many studies on antioxidant supplements do not show a health benefit is because antioxidants tend to work best in combination with other nutrients, plant chemicals, and even other antioxidants.
Explanation: