Explanation:
<u>The first one is a base</u>
<u>The second one is an acid</u>
A base has a pH above 7
An acid has a pH below 7
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The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.
The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.
<h3>Conservation law</h3><h3 />
The given equation has been assessed as follows:
The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.
The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.
The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.
The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.
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Decomposition is a chemical reaction that breaks the reactant into two or more products. Moles of nitrogen gas
in the cylinder is 1.63 moles.
<h3>What is the ideal gas equation?</h3>
The ideal gas equation states the relation of the hypothetical ideal gas according to the pressure, volume, temperature and moles of the gas. It is given by,

Where,
Pressure (P) = 2000 kPa
Volume (V) = 2L
Temperature (T) = 295 K
Gas constant (R)= 0.08206
Substituting values in the equation:

Therefore, 1.63 moles are produced.
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Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2
<span>B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) </span>
<span>C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) </span>
<span>D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
</span>Answer is: B.
Chemical reaction: F<span>e</span>₂O₃<span>(s) + 3CO(g) → 2Fe(s) + 3CO</span>₂<span>(g);</span>ΔH = <span>+ 26.3 kJ.
When one mole of iron(III) oxide reacts 26,3 kJ of energy is required and for 3,2 moles of iron(III) oxide 3,2 times more energy is required.</span>
<u>Answer:</u> The equilibrium concentration of
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of
= 2.00 M
The given chemical equation follows:

<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of
is 0.332 M