You have many different ways possible. If you have a freezer in the classroom you could do that. If you made it really cold in the room and stopped all air flow in the room so it stayed cold you would be able to freeze it.
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Answer:
This is the change of one type of matter into another type of matter
Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
<span>The Keq expression for I2(s) + H2O(l) H + (aq) + I-(aq) + HOI(aq) equilibrium is
H2(g) + I2(g) <----> 2 HI(g).In this Keq expression hydrogen reacts with iodine gives Hydrogen Iodide.HI is a diatomic molecule and it is one of the primary source of iodine which act as a reducing agent.Totally it is the reverse chemical equilibrium reaction.</span>
Answer:
ΔHrxn = 178.3 kJ/mol
Explanation:
Using Hess's law, you can obtain ΔHrxn from ΔHf of products and reactants, thus:
<em>Hess's law: </em>
ΔHrxn = <em>∑nΔHf products - ∑nΔHf reactants</em>
<em>Where n are moles of reaction</em>
<em> </em>
Thus, from the reaction:
CaCO₃(s) → CaO(s) + CO₂(g)
ΔHrxn = <em>ΔHf </em>CaO(s)<em> + ΔHf </em>CO₂(g)<em> - ΔHf </em>CaCO₃(s)
ΔHrxn = -635.1kJ/mol + (-393.5kJ/mol) - (-1206.9kJ/mol)
<em>ΔHrxn = 178.3 kJ/mol</em>
