18 moles of water are produced in the above reaction.
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Answer:
The correct answer is 1.21 L.
Explanation:
Based on the given information, the reaction will be,
CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)
By using the standard values of the substances, the standard enthalpy of the reaction is,
ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol
= -285.3 kJ/mol
The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.
Now the number of moles of chlorine needed to react to produce 5.00 kJ is,
= 5.00 kJ × 3 mol Cl2/285.3 kJ
= 0.0526 mol Cl2
Now the volume of chlorine gas at 27degree C and 812 mmHg will be,
Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm
= 1.21 L
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
The answer is 0.975 L
Volume = mol/Molarity
We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:
Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol
So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles
Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L
