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Answer:
0.46M NaS₂O₃ (Assuming KIO₃ solution with a concentration of 1.0M)
Explanation:
Based on the reaction:
6 Na₂S₂O₃ + KIO₃ + 5 KI + 3 H₂SO₄ → 3 Na₂S₄O₆ + 3 H₂O + 3 K₂SO₄ + 6 NaI
<em>6 moles of Na₂S₂O₃ react per mole of KIO₃</em>
Assuming the molarity of the KIO₃ solution is 0,1M:
Moles of KIO₃: = 5.0x10⁻³L ₓ (0.1 mol / L) = <em>5.0x10⁻⁴ moles</em>
As 6 moles of thiosulfate reacted per mole of iodate:
5.0x10⁻⁴ moles KIO₃ ₓ (6 moles Na₂S₂O₃ / 1 mole KIO₃) =
<em>3.0x10⁻³ moles of Na₂S₂O₃. </em>In 6.5mL (6.5x10⁻³L):
3.0x10⁻³moles Na₂S₂O₃ / 6.5x10⁻³ L = 0.46M NaS₂O₃
Answer: - 986.6 kj/mol
Explanation:
1) Equation given:
CaO(s) + H₂O (l) → Ca (OH)₂ (s) δh⁰ = −65.2 kj/mol
2) Standard enthalpies of formation given:
CaO, δhf⁰ = −635.6 kj/mol
H₂O, δhf⁰ = −285.8 kj/mol
3) Calculate the standard enthalpy of formation of Ca(OH)₂.
δh⁰ = ∑δh⁰f of products - ∑ δh⁰f of reactants
Using the mole coefficients of the balanced chemical equation:
δh⁰ = δh⁰f Ca(OH)₂ - δh⁰f CaO - δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = δh⁰ + δh⁰f CaO + δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = - 65.2 kj/mol − 635.6 kj/mol) − 285.8 kj/mol) = - 986.6 kj/mol.
Answer:
M = 3.69 M.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the molar concentration of the 1.29 moles of KCl in 350 mL of solution by recalling the mathematical definition of molarity as the division of the moles by the volume in liters, in this case 0.350 L; thus, we proceed as follows:
Which gives molar units, M, or just mol/L.
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Answer:
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Explanation:is this what you meant?
