<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

8 0

3 years ago

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts

Y_Kistochka [10]

Answer:

Explanation:

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?

Mg + O2 → MgO (unbalanced)

first, balance the equation

2Mg +O2-------> 2MgO

two magnesium atoms react with one diatomic oxygen molecule

there is a 1:1 ratio of magnesium to oxygen atoms

but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms

the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.

6 0

3 years ago