The molarity of the potassium acetate solution given the data is 1.584 M
<h3>What is molarity? </h3>
This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
<h3>How to determine the mole of CH₃COOK</h3>
- Mass of CH₃COOK = 19.4 g
- Molar mass of CH₃COOK = 98 g/mol
- Mole of CH₃COOK =?
Mole = mass / molar mass
Mole of CH₃COOK = 19.4 / 98
Mole of CH₃COOK = 0.198 mole
<h3>How to determine the molarity of CH₃COOK</h3>
- Mole of CH₃COOK = 0.198 mole
- Volume = 125 mL = 125 / 1000 = 0.125 L
- Molarity of CH₃COOK = ?
Molarity = mole / Volume
Molarity of CH₃COOK = 0.198 / 0.125
Molarity of CH₃COOK = 1.584 M
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2SO₂ + O₂ = 2SO₃
n(O₂)=1 mol
n(SO₂)=2n(O₂)
n(SO₂)=2 mol
Answer:
NiCO3 (s) + 2H+ (aq) → H2O (l) + CO2 (g) + Ni2+ (aq)
Explanation:
To write the complete ionic equation:
1. Start with a balanced molecular equation.
2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
3. indicate the correct formula and charge of each ion
4. indicate the correct number of each ion
5. write (aq) after each ion
6. Bring down all compounds with (s), (l), or (g) unchanged.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
Answer:
NO3 that is the answer to the question
