Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.
<h3>How can iron (iii) chloride be formed from iron fillings?</h3>
Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:
- Iron fillings + Cl₂ → FeCl₃
Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.
In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.
Learn more about iron (iii) chloride at: brainly.com/question/14653649
#SPJ1
Answer: 15062.4 Joules
Explanation:
The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of food = 200.0g
C = 4.184 j/g°C
Φ = (Final temperature - Initial temperature)
= 83.0°C - 65.0°C = 18°C
Then, Q = MCΦ
Q = 200.0g x 4.184 j/g°C x 18°C
Q = 15062.4 J
Thus, 15062.4 joules of heat energy was contained in the food.
Answer:
1.20atm
Explanation:
Given parameters:
Partial pressure of gas 1 = 0.35atm
Partial pressure of gas 2 = 0.20atm
Partial pressure of gas 3 = 0.65atm
Unknown:
Total pressure of the gas mixture = ?
Solution:
To solve this problem, we need to recall and understand the Dalton's law of partial pressure.
Dalton's law of partial pressure states that "the total pressure of a mixture of gases is equal to the sum of the partial pressure of the constituent gases".
Total pressure =Pressure of gas(1 + 2 + 3)
The partial pressure is the pressure a gas would exert if it alone occupied the volume of the gas mixture.
Now we substitute;
Total pressure = (0.35 + 0.20 + 0.65)atm = 1.20atm
Answer:
The law that suggests that at a constant pressure and the volume of gas directly proportional to its temperature is the Boyle's law.
Explanation:
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
