The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Possibly wet and unstable
1) Carbon dioxide is a gas, so when 
is evolved in the reaction, it appears as bubbles. The gas released extinguishes the fire and it can turn lime water milky.
2) When 
is released in a decomposition reaction we can identify by the strong pungent smell of the gas released.
3) Saturated citric acid can cause corrosion of the metal layers present in the pipes. So, before draining out any acid it is neutralized so that the pipes and other plumbing works do not get damaged leading to leaks in the drainage system.
SAMPLE A - <span>pure substance.
</span>SAMPLE B - <span>homogeneous mixture.
</span>SAMPLE C - <span>heterogeneous mixture.
</span>Pure substance - <span>constant composition and properties.</span>
Homogeneous mixture - same uniform appearance and composition.
Heterogeneous mixture - <span>not </span>uniform<span> in composition, two phases (liquid and dust).
</span>
Answer:
The ring of fire
Explanation:
The ring of fire is where they meet or at the fault lines
