Answer:
Different types of hot or cold items can be stored in a thermos and power cannot enter or exit the system when the thermos lid is tightly closed
Explanation:
Closed systems are those that do not interact or do not exchange energy with the environment that surrounds them, that is why internal temperatures and conditions are maintained.
The human body is an open system, that is, it would be the opposite of the thermos since we constantly exchange energy with the environment through sweating, emission of gases, urine, feces, and the ingestion of food.
Thermoses are systems specially created to maintain a medium, it will be maintained if its lid is hermetically closed to prevent heat leakage or entry in situations of cold fluids.
Answer:
0.01836 M
Explanation:
Again the reaction equation is;
Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)
E°cell= 0.77 V
Ecell= 0.78 V
[Mn2+] = 0.040 M
[Fe2+] = the unknown
n=2
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]
0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]
0.01/ -0.0296= log [Fe2+] /[0.040]
-0.3378= log [Fe2+] /[0.040]
Antilog(-0.3378) = [Fe2+] /[0.040]
0.459= [Fe2+] /[0.040]
[Fe2+] = 0.459 × 0.040
[Fe2+] = 0.01836 M
Answer:
b)15.0°C
Explanation:
Specific Heat of Water=4.2 J/g°C
This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.
∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.
80×4.2 J=336 J
Total Energy Provided=1680 J
The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}
Temperature increase=
=5°C
Initial Temperature =10°C
Final Temperature=Initial + Increase in Temperature
=10+5=15°C
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
