Answer:

Explanation:
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)
ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B1%28-1675.7%29%20%2B%202%280%29%5D%20-%20%5B2%280%29%20-%201%28-824.3%29%5D%5C%5C%26%20%3D%20%26%20-1675.7%20%2B%20824.3%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20change%20is%20%7D%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%7D)
Answer:
Tungsten is used for this experiment
Explanation:
This is a Thermal - equilibrium situation. we can use the equation.
Loss of Heat of the Metal = Gain of Heat by the Water

Q = mΔT
Q = heat
m = mass
ΔT = T₂ - T₁
T₂ = final temperature
T₁ = Initial temperature
Cp = Specific heat capacity
<u>Metal</u>
m = 83.8 g
T₂ = 50⁰C
T₁ = 600⁰C
Cp = 
<u>Water</u>
m = 75 g
T₂ = 50⁰C
T₁ = 30⁰C
Cp = 4.184 j.g⁻¹.⁰c⁻¹

⇒ - 83.8 x
x (50 - 600) = 75 x 4.184 x (50 - 30)
⇒
=
j.g⁻¹.⁰c⁻¹
We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹
So metal Tungsten used in this experiment
Answer: the false statement is:
The only functional group possible in a hydrocarbon is the double bond
Answer:
Double replacement reaction.
Explanation:
The Na and Ag atoms both (double) trade places (replacement) with each other.