Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
Answer:
Keq =1.50108
Explanation:
The given reactionis
C₂H₂(g) +2H₂(g) -------------> C₂H₂(g)
ΔG0 f=ΔG0f n (products) - ΔG0f n (reactants )
= -32.89 kJ/mol - (209.2 kJ/mol+2*0.0 kJ/mol)
= - 242.09kJ/mol
ΔG= -RTlnKeq
ln Keq = -ΔG/RT
=-(- 242.09kJ/mol ) / 2 k cal /mol*298 K
=0.406
Keq =e0.406
Keq =1.50108
Exothermic reactions is where energy is released. Exothermic reactions are reactions that release energy into the environment in the form of heat. Exothermic reactions feel warm or hot or may even be explosive.
