FeO + Al -> AlO + Fe
As you can see, this type of reaction would lead to be a Single Placement (or Replacement) reaction!
Hope this helps! Best wishes!
Answer:
the answer is b hope it helped
Explanation:
The unsaturated zone is the portion of the subsurface above the groundwater table. The soil and rock in this zone contains air as well as water in its pores. ... Unlike the aquifers of the saturated zone below, the unsaturated zone is not a source of readily available water for human consumption
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
50,849.25 Joules
Explanation:
The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:
Q = mcΔT, where ΔT represents the change in temperature.
In the case of the iron block:
m = 75 g
c = 0.449 J/g °C
ΔT = 1535 - 25 = 1510 °C
Therefore,
Q = 75 g x 0.449 J/g °C x 1510 °C
= 50,849.25 Joules
<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>
