Answer:
D. Two reactants combine to form one product
Explanation:
The definition of synthesis in chemistry is "the production of chemical compounds by reaction from simpler materials."
A is decomposition, B is DOUBLE replacement, and C is SINGLE replacement.
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
<u>Answer:</u>
<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ
<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ
<u>Explanation:</u>
Relationship between standard Gibbs free energy and standard electrode potential follows:
............(1)
The given chemical equation follows:

<u>Oxidation half reaction:</u>
( × 2)
<u>Reduction half reaction:</u> 
We are given:

Putting values in equation 1, we get:

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ
The given chemical equation follows:

<u>Oxidation half reaction:</u>
( × 6)
<u>Reduction half reaction:</u> 
We are given:

Putting values in equation 1, we get:

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ
Answer:
Friston to energy
Explanation:
Friction to Energy as its a energy stooping the object form moving