Answer:
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
Which can be written in terms of x, according to the ICE table:
Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
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<h3>
Answer:</h3>
2265 g Fe₃O₄
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O
[Given] 705.0 g H₂O
<u>Step 2: Identify Conversions</u>
[RxN] 4 mol H₂O → 1 mol Fe₃O₄
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Fe - 55.85 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol
<u>Step 3: Convert</u>
- Set up stoich:
- Multiply/Divide/Cancel units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄
It can be explained in<span> simple </span>terms<span>: Oxidation is the loss of electrons or an increase </span>in<span> oxidation state by a molecule, atom, or ion. </span>Reduction<span> is the gain of electrons or a decrease </span>in<span> oxidation state by a molecule, atom, or ion.</span>
Answer : The mass of potassium phosphate needed will be 3.98 grams.
Explanation : Given,
Molarity of potassium phosphate = 0.75 M = 0.75 mole/L
Molar mass of potassium phosphate = 212.27 g/mole
Volume of solution = 25.0 mL = 0.025 L (conversion used : 1 L = 1000 mL)
Molarity : It is defined as the number of moles of solute present in one liter of solution.
In this problem solute is, potassium phosphate.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the mass of potassium phosphate needed will be 3.98 grams.