Chemistry, 12.07.2019 04:30 xaviiaquino3378

1. describe how the rube goldberg device follows the law of conservation energy. be sure to include both parts of the law of conservation of energy in your description. will give crown thingay

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Rube Goldberg machines, follows the “law of conservation of energy”, this machine uses a chain reaction to complete a simple task in different manner,

In this machine, a various instant kinetic energy is transferred into potential energy and vice-versa , so the total energy remain conserved. Thus Rube Goldberg machines follows law of conservation of energy.

In this machine when marble rolls down from top it has kinetic energy. Then the marble hits and falls into a plastic cup, its kinetic energy is transferred into potential energy that lifts the soda can.Then potential energy is changed into kinetic energy. When the marble make contact with the next marble, then its kinetic energy is transferred to other marble. When marble falls its gravitational energy is released. This gravitational energy built up into kinetic energy, which was transferred to the domino. Each domino passes its energy to the next in a chain reaction until the last one sets off the mousetrap. As the mousetrap is set off, it discharges all of its stored potential energy which pulls back the scissors. The string tying down the balloon is cut, freeing the balloon and its potential energy stored.

Answer from: Quest

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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1. describe how the rube goldberg device follows the law of conservation energy. be sure to include...