Two isentropic processes in an otto cycle are replaced with polytrophic processes. given: g1: n=1.3 g2: r=8 g3: p.=95 kpa g4: t,=15°c g5: ts=1200°c determine: d1: the heat transferred to the cycle d2: the heat rejected by the cycle d3: thermal efficiency

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

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answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol