Answer:
Robot 1 : 150 J
Robot 2: 40 J
Explanation:
30 N x 5m = 150 J
20 N x 2m = 40 J
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We tend to think the greater the mass, the greater the volume but that is not true. What weighs more a pound of lead or a pound of feathers? Obviously they are the same however there would be a lot more feathers than lead. Mass and volume are not really related.
Answer:
moles of ammonia produced = 0.28 moles
Explanation:
The reaction is

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia
So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen
= 0.42 moles of hydrogen molecule.
this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia
the moles of ammonia produced = 0.28 moles
Here the nitrogen is limiting reagent.
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction
.

![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Is there choices to this question? cant answer it without choices