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
The answers is 67 don’t ask me I don’t know but trying to get free points
1 kg/L ----------- 0.001 kg/mL
22.4 kg/L ------- ??
22.4 x 0.001 / 1 => 0.0224 kg/mL
The answer is 3. As 5 * 3 = 15.
Is there any other equations I may be able to help you with? :)
Answer:
75.15 mol.
Explanation:
- Firstly, we need to write the balanced equation of the reaction:
<em>Fe₂O₃ + 3CO → 2Fe + 3CO₂.</em>
It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.
∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.
- we need to calculate the no. of moles of (4000 g) of Fe₂O₃:
<em>no. of moles of Fe₂O₃ = mass/molar mass</em> = (4000 g)/(159.69 g/mol) = <em>25.05 mol.</em>
<u>Using cross multiplication:</u>
1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,
∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.
<em>∴ The no. of moles of CO needed</em> = (3.0 mol)(25.05 mol)/(1.0 mol) =<em> 75.15 mol.</em>
