Answer:
Heat flux = 13.92 W/m2
Rate of heat transfer throug the 3m x 3m sheet = 125.28 W
The thermal resistance of the 3x3m sheet is 0.0958 K/W
Explanation:
The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

The heat flux can be defined as the amount of heat flow by unit of area.
Using the previous calculation, we can estimate the heat flux:

It can also be calculated as:

The thermal resistance can be expressed as

For the 3m x 3m sheet, the thermal resistance is

Answer:
acetic acid, sodium hydroxide
Explanation:
A strong acid is an acid that ionizes in water to give all its hydrogen ion. Weak acid only ionize to a certain degree. Acetic acid (CH3COOH) only ionize to give one hydrogen ion despite having other hydrogen atom. This account for its weak nature as an acid as shown below:
CH3COOH <=> H^+ + CH3COO^-
A strong base is a base that ionizes in water to give all it hydroxide ion. Sodium hydroxide(NaOH) ionizes to give all its hydroxide ions. This make it a strong base as shown below;
NaOH <=> Na^+ + OH^-
I want to say 59 atomic mass units<span> is beta decay, but I'm just guessing from previous Chem. class.</span>
<span>Wave A will have a higher pitch than wave B.. This is not true becasue b has the higher PITCH becasue it's closer together.
Wave B will have a lower pitch than wave A... This basically A but worded differently.
Wave A will have a louder sound than wave B... This is CORRECT becasue it's louder the waver are bigger, so it's louder. But it has a lower pitch, 2 different things.
Wave B will have a louder sound than wave A. False..</span><span />
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium constant is 
Explanation:
From the question we are told that
The chemical reaction equation is

The voume of the misture is
The molar mass of
is a constant with value of 
The molar mass of
is a constant with value of 
The molar mass of
is a constant with value of 
Generally the number of moles is mathematically given as

For 


For 


For 


Generally the concentration of a compound is mathematicallyrepresented as

For 
![Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}](https://tex.z-dn.net/?f=Concentration%5BFe_2%20O_3%5D%20%3D%20%5Cfrac%7B0.222125%7D%7B5.4%7D)
For 
![Concentration[H_2] = \frac{1.815}{5.4}](https://tex.z-dn.net/?f=Concentration%5BH_2%5D%20%3D%20%5Cfrac%7B1.815%7D%7B5.4%7D)

For 
![Concentration [H_2O] = \frac{0.12}{5.4}](https://tex.z-dn.net/?f=Concentration%20%5BH_2O%5D%20%3D%20%5Cfrac%7B0.12%7D%7B5.4%7D)

The equilibrium constant is mathematically represented as
![K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5Bconcentration%20%5C%20of%20%5C%20product%5D%7D%7B%5Bconcentration%20%5C%20of%20%5C%20reactant%20%5D%7D)
Considering 
And 
At equilibrium the

