Acetylene burns in air according to the following equation: c2h2(g) + 5 2 o2(g) → 2 co2(g) + h2o(g) δh o rxn = −1255.8 kj given

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Acetylene burns in air according to the following equation: c2h2(g) + 5 2 o2(g) → 2 co2(g) + h2o(g) δh o rxn = −1255.8 kj given δh o f of co2(g) = −393.5 kj/mol and δh o f of h2o(g) = −241.8 kj/mol, find δh o f of c2h2(g).

Hermina Ferry · Accepted Answer

-227 kJExplanation:The balanced chemical reaction is, The expression for enthalpy change is,where,n = number of moles (as heat of formation of substances in their standard state is zeroNow put all the given values in this expression, we getTherefore, the enthalpy change for is -227 kJ.

Hermina Ferry · Answer

its going to be mutualism cuz they both benefit from the relationship.~batmans wife dun dun

Hermina Ferry · Answer

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.the first pair: ne (neon) and kr (krypton), are noble gases found in group 8the second pair: li (lithium) and k (potassium), found in group 1the third pair: s (sulfur) and te (tellurium), found in group 6

Jakob Littel

Acetylene burns in air according to the following equation: c2h2(g) + 5 2 o2(g) → 2 co2(g) + h2o(g) δh o rxn = −1255.8 kj given

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