because it can influence how frequently and sufficiently the particles collide depending on the space it has to do so, for example a large surface area would be have a slower rate of reaction and a lower temperature. (the rate of reaction in terms of concentration, it is diffused from high to low)
Answer: The fourth material that is added to the blast furnace is HOT AIR which provides OXYGEN for used for combustion of carbon (Coke).
Explanation:
Iron is the second most abundant metal found in the earth's crust after aluminium. It is not found in the free metallic state but are extracted from rocks which are rich in iron that contains other materials. These are known are iron ores and the most common iron ores are haematite ( Fe2O3).
Iron can be extracted from its ore with the used of blast furnace. The materials used for extraction of iron includes:
--> Coke
--> haematite( iron ore)
--> limestone and
--> Hot air.
The iron ore is first roasted in air so that iron(III)oxide is produced. The iron(III)oxide is then mixed with coke and limestone and heated to a very high temperature. Hot air is introduced into it from the bottom of the furnace. The coke is oxidizes the the oxygen in the hot air blast to liberate carbondioxide.
Answer:
When you put sugar inside of a cup with water, the sugar is still visible because it's molecules have just gotten in touch with water molecules. The sugar molecules are still attracted to each other but as you stir it, it seems to disappear but not completely. When the water is stirred sugar mix with water and water molecules place themselves between the sugar ones.
Conclusion: It suggests that the sugar molecules are more attracted to water molecules which is why they easily separate from each other.
Answer:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ