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
We all know, Density = Mass/ volume.
When volume increases and the mass remains the same , the density will decrease considerably.
From the reaction between Cu and HNO₃, the formed gas is NO₂ instead of NO₃. Hence the correct balanced equation would be,
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O<span>(l)
Here, Cu goes to </span>Cu(NO₃)₂ by changing its oxidation number from 0 to +2 while NO₃⁻ goes to NO₂ by reducing its oxidation state from +5 to +4 . Hence Cu is oxidized by HNO₃ in the reaction.
Answer:
Fe2O3 + __3_H2O --> __2_Fe(OH)3
Explanation:
____Fe2O3 + __3_H2O --> __2_Fe(OH)3
Fe -2 Fe -2
H - 6 H - 6
O - 6 O- 6
