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Answer:
588.2 mL
Explanation:
- FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)
First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):
- moles = molarity * volume
- 187 mL * 0.692 M = 129.404 mmol Fe⁺²
Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:
- 129.404 mmol Fe⁺² *
= 258.808 mmol KOH
Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:
- volume = moles / molarity
- 258.808 mmol KOH / 0.440 M = 588.2 mL
Answer:
V₂ =279.9 cm³
Explanation:
Given data:
Initial volume = 360 cm³
Initial temperature = 50°C
Initial pressure = 700 mmHg
Final volume = ?
Final temperature = 273 k
Final pressure = 1 atm
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Solution:
<em>We will convert the mmHg to atm.</em>
700/760 = 0.92 atm
<em>and °C to kelvin.</em>
50+273 = 323 K
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm
V₂ = 290417.6 atm .cm³. K / 323 k. atm
V₂ =279.9 cm³
Is there any answer choices before i get started on working out the problem