-2.5089 kJ is the amount of heat is released by the combustion of the naphthalene.
<h3>How we calculate the released heat?</h3>
Released amount of heat is calculated by using the below formula:
Q = -mcΔT, where
m = mass of naphthalene = 0.25g
c = specific heat of calorimeter = 4.13 kJ/°C
ΔT = change in temperature = 26.43°C - 24°C = 2.43°C
Putting all these values in the above equation, we get
Q = - (0.25 × 4.13 × 2.43) = - 2.5089 kJ
Hence, - 2.5089 kJ heat is released.
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brainly.com/question/13439286
The balanced equation for the neutralisation reaction is as follows
Ca(OH)₂ + H₂SO₄ ---> CaSO₄ + 2H₂O
stoichiometry of Ca(OH)₂ to H₂SO₄ is 1:1
equivalent number of acid reacts with base
number of H₂SO₄ mol reacting - 2 mol
according to molar ratio of 1:1
number of Ca(OH)₂ mol = number of H₂SO₄ moles
therefore number of Ca(OH)₂ moles required - 2 mol
Answer:
8
Explanation:
look at the periodic table find sulfur then count what level is it on by rows
It was probably because there was no organization to the order of the elements so they got frustrated since they didn't know where things went.
Answer:
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
Explanation:
This is an easy redox reaction:
Br₂ + 2KI → I₂ + 2KBr
We determine the oxidation states.
0 for the elements at ground state.
K does not change the oxidation state during the reaction.
Bromine is reduced to bromide (oxidation state decreases)
and iodide is oxidized to Iodine (oxidation state increases)
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
In oxidation, electrons are released while in reduction, the electrons are gained. To make the ionic equation, we just add K⁺
So we sum both reactions
Br₂ + 2e⁻ + 2I⁻ + 2K⁺ ⇄ 2e⁻ + I₂ + 2Br⁻ + 2K⁺
We cancel the electrons on both sides of the equation:
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺