Answer:
Final Temperature = 28.2 oC
Explanation:
Information given;
Mass of Iron = 20.8g
Initial Temperature of Iron = 100C
Mass of water = 55.3g
Initial temperature of water = 25.3 C
The presence of a coffee cup calorimeter hints that there is no heat loss to the surrounding and that the iron and water are at thermal equilibrium.
Thermal equilibrium means that there is no heat transfer going on between the bodies, which simply means that the bodies are at the same temperature.
Hence, both bodies would the same final temperature (T2)
H = M * C * ΔT (For iron)
H = 20.8 * 0.449 * ( 100 - T2)
H = 9.3392 ( 100 - T2)
H = 933.92 - 9.3392T2
H = M * C * ΔT (For water)
H = 55.3 * 4.184 * (T2 - 25.3)
H = 231.3752 (T2 - 25.3)
H = 231.3752T2 - 5853.79
Since they are in thermal equilibrium it means H (Iron) = H (water).
This leads to;
933.92 - 9.3392T2 = 231.3752T2 - 5853.79
231.3752T2 + 9.3392T2 = 5853.79 + 933.92
240.7144 T2 = 6787.71
T2 = 28.2 oC
Answer:
1) The temperature inside a diesel engine is high enough for two atmospheric gases nitrogen and oxygen to react and combine to form nitrogen oxides, NOₓ
At very high temperatures of above 1,300 °C, which are temperatures obtainable in the internal combusting chamber of a diesel engine under 100% load, air containing N₂ and O₂ is sucked in, and they combine to produce approximately 85% of the NOₓ pollutants produced from moving engine sources
2) a) N₂O contributes to the Greenhouse effect and therefore global warming
b) NO₂ contributes to the formation of smog and NO₂ combines with water to produce nitric acid which falls as acid rain, and causes irritation in the eyes and the respiratory system airways
c) Nitric oxide gas, NO, has the effect of depleting the Ozone layer which increases global warming and increased penetration of UVB
Explanation:
<span>Ag+(aq) + e– → Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V
=>
</span>
<span>2Ag+(aq) + 2e– → 2Ag(s) Eo = +0.80 V
Pb(s) → Pb2+(aq) + 2e– - Eo = – (-0.13 V) = +0.13 V
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2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)
</span>
<span>E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.
Answer: + 0.93V
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