Answer:
The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.
Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.
Explanation:
CH4(g) + 2H2O(g) ⇆ CO2(g) + 4H2(g)
CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0
ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0
S°(J/K·mol): 186.1 188.8 213.7 130.7
ΔG<0 to be spontaneous
ΔG = ΔH- TΔS <0
ΔH = ∑nΔH(products) - ∑nΔH(reactant)
ΔH = (-393.5) - (–74.87 + 2*–241.8)
ΔH = 164.97 kJ = 164970 J
ΔS = ∑nΔS(products) - ∑nΔS(reactant)
ΔS = (213.7 + 4*130.7) - (186.1 + 2*188.8)
ΔS = 172.8 J
0 > 164970 J - T* 172.8 J
-164970 J > - T* 172.8 J
954.7< T
The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.
Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.
Sorry but I only know a few
17- decomposers
18- decomposition
21- 10% is lost so 90%
Answer:
C) H⁺
Explanation:
When we are balancing the reaction<em> in an acid medium</em>, hydrogen is balanced using the <u>H⁺</u> species. This is most likely the intended answer of your question.
When the reaction takes place not in an acid medium, but in <em>an alkaline one</em>, then hydrogen is present as the OH⁻ species. However this option is not given in your question.
Thus the answer is option C).
Answer:
sorry
Explanation:
I don't know the answer this is really confusing but I am really sorry you have to do this.
Answer:
0.297 mol/L
Explanation:
<em>A chemist prepares a solution of potassium dichromate by measuring out 13.1 g of potassium dichromate into a 150 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium dichromate solution. Be sure your answer has the correct number of significant digits.</em>
<em />
Step 1: Calculate the moles corresponding to 13.1 g of potassium dichromate
The molar mass of potassium dichromate is 294.19 g/mol.
13.1 g × (1 mol/294.19 g) = 0.0445 mol
Step 2: Convert the volume of solution to L
We will use the relationship 1 L = 1000 mL.
150 mL × (1 L/1000 mL) = 0.150 L
Step 3: Calculate the concentration of the solution in mol/L
C = 0.0445 mol/0.150 L = 0.297 mol/L
