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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The combustion reaction is as expressed,
CxHy + O2 --> CO2 + H2O
The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.
Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.
moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14
The empirical formula for the hydrocarbon is therefore, CH₂.
I think the answer is B. the sum of the enthalpy changes of the intermidiate reactions
4. 2Li + 2H2O -> 2LiOH + H2
5. C6H12O6 + 6O2 -> 6CO2 + 6H2O
6. Zn + 2HCl -> ZnCl2 + H2
9. H2SO4 + Pb -> PbSO4 + H2
10. Ca(OH)2 + NH4Cl -> NH4 + CaCl2 + H2O
thats all i know
