Answer:
The amount of salt in the tank at the end of an additional 10 minutes are 4.38lb.
Explanation:
<u>Situation 1:</u>
Tank with 100 gallons of fresh water
<u>Situation 2:</u>
Tank with 100 gallons of fresh water + water with 0.5lb of salt per gallon
After 10 minutes, as the rate in which the new water is poured is 2 gallons per minute, the result is 20 gallons added (2×10=20) . And taking in account that the water contains 0.5 lb of salt per gallon the amount of salt added is 20×0.5= 10lb of salt.
That amount of salt is now in all the water inside the tank which is 100 gallons+ 20 gallons= 120 gallons. <em>That means that in situation 2 we have 10lb of salt in 120 gallons of water.</em>
That mixture is allowed to leave the tank at a rate of 2 gallons per minute so we will have after 10 minutes: 120 gallons- (2×10) gallons= 100 gallons remaining in the tank. And the amount of salt if we remember that we had 10lb in 120 gallons, now in 100 gallons we will have: (100 gallons × 10lb of salt)/ 120 gallons= 8.33 lb of salt.
<u>Situation 3:</u>
Tank with 100 gallons of water with 8.33lb of salt.
After 10 minutes in which fresh water is poured in the tank at a rate of 9 gallons per minute, the result is: 9×10= 90 gallons added to the tank. So now we have 100+90=190 gallons of water in the tank. <em>That means in situation 3 we have 8.33 lb of salt in 190 gallons of water. </em>
That mixture is leaving the tank at a rate of 9 gallons per minute so we have after 10 minutes: (190- (9×10))= 100 gallons of mixture remaining in the tank.
And the amount of salt if we remember that we had 8.33lb in 190 gallons, now in 100 gallons we will have: (100 gallons × 8.33lb of salt)/ 190 gallons= 4.38 lb of salt.
Answer:
Explanation:The equation for this is F = -GmM/R^2 where the minus sign says the force is attractive m is 10 kg, M is 20 kg and R is 5 meters. If you crunch the numbers you get an answer of:
Answer: 600
I hope this helps you
When the particles of a substance (usually a liquid) is heated up, its particles absorb the energy provided thereby increasing their kinetic energy resulting to more movement of the individual particles.
<u>Answer:</u> The value of 
for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K
<u>Explanation:</u>
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20S%5Eo_%7B%28NO_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28N_2%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20S%5Eo_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28240.06%29%29%5D-%5B%281%5Ctimes%20%28191.61%29%29%2B%282%5Ctimes%20%28205.14%29%29%5D%5C%5C%5C%5C%5CDelta%20S%5Eo_%7Brxn%7D%3D-121.77J%2FK)
Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K
We are given:
Moles of nitrogen gas reacted = 1.90 moles
By Stoichiometry of the reaction:
When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K
So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = 
Hence, the value of for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K
