Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.
(a) Barium nitrate is a <em>salt</em> formed by the <em>cation</em> barium Ba²⁺ and the <em>anion</em> nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a <em>salt</em> formed by the <em>cation</em> potassium K⁺ and the <em>anion</em> chlorate ClO₃⁻. Its formula is KClO₃.
(b) The balanced equation for the decomposition of potassium chloride is:
2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)
(c) The balanced equation for the decomposition of barium nitrate is:
Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)
(d) The balanced equations of metals with oxygen to form metal oxides are:
- 2 Mg(s) + O₂(g) ⇄ 2 MgO(s)
- 4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)
- 4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)
im missing something though
Answer:
<em>K</em><em>+</em><em>Cl</em><em /><em>KCl</em>
Explanation:
because the reaction is between metal Potassium and Non-metal Chlorine
The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
To learn more about enthalpy here
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Answer:
Counting the number of colonies that arise on a pour plate can calculate the concentration by multiplying the count by the volume spread on the pour plate. Direct counting methods are easy to perform and do not require highly specialized equipment, but are often slower than other methods
Explanation:
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