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AleksAgata [21]
3 years ago
8

The circles in the diagrams below represent energy levels in an atom, and the arrows show electron (blue dot) transitions from o

ne energy level to another. (the spacing between circles represents differences in energy: a larger spacing means a greater difference in energy.) assuming that the transitions occur as photons are emitted, rank the atoms based on the photon energy, from highest to lowest. hints

Chemistry
1 answer:
Hoochie [10]3 years ago
6 0
Ranking of the atom from highest to lowest is as follows:
Highest
Arrow = from outer edge to center
2nd Highest
Arrow = second closest ring to the outer edge to center
3rd Highest
Arrow = middle circle to center
Lowest
Arrow = outer edge to middle circle

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What is the concentration of an hbr solution if 12.0 ml of the solution is neutralized by 15.0 ml of a 0.25 m koh solution?
Snowcat [4.5K]
The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration 
number of KOH moles reacted = number of HBr moles reacted 
number of HBr moles reacted - 0.00375 mol 
if 12 mL of HBr contains - 0.00375 mol 
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
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3 years ago
Helooo i need help please :)
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A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

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