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
(NO3)2+H2O+NO that is the answer
Answer and explanation:
A) An ideal fuel must:
- easy to transport and storage.
- have a high calorific value.
B) The <em>calorific value</em> for a fuel is the amount of heat - measured in Joules- which is produced during the complete combustion of the fuel. It is expressed in Joules per Kg of fuel (J/kg).
C) From the data:
mass of fuel = 2 kg
heat produced = 48,000 KJ
We calculate the calorific value by dividing the heat produced by the mass of fuel, as follows:
calorific value = heat produced/mass of fuel = (48,000 KJ)/(2 kg)= 24,000 kJ/kg
Since 1 KJ= 1000 J, we can express the calorific value in J/kg as follows:
24,000 kJ/kg x 1000 J/1 kJ = 2.4 x 10⁷ J/kg
Ozone 03 have 2 sigma bond and 1 pie bond.
just remember this and you can do any problem for pie and sigma with this.
single bond = 1 sigma
double bond = 1 sigma and 1 pie
triple bond = 1 signa and 2 pie
The relationship between the evaporation rate and the likelihood that the liquid will be flammable is its flash point number (or the lowest temperature for which the liquid ignites) and its activation energy (how much energy is required to ignite it). With volatile liquids or gases these numbers seem to be lower.
