Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant ![K_c= \dfrac{[CO][H_2]}{[H_2O]}](https://tex.z-dn.net/?f=K_c%3D%20%20%5Cdfrac%7B%5BCO%5D%5BH_2%5D%7D%7B%5BH_2O%5D%7D)
The equilibrium constant 
The equilibrium constant 
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :
Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
![0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}](https://tex.z-dn.net/?f=0.03905%20%3D%20%20%5Cdfrac%7B%5B0.17%2Bx%5D%5Bx%5D%7D%7B%5B0.91%20-x%5D%7D)
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095
x -x²
x² - 0.13095
x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
Hello
the answer is 43.129310000000004
Have a nice day
I don't know what the problem is, but here are some rues to help you out:
- All non-zero figures are significant
- When a zero falls between non-zero digits, that zero is significant.
- When a zero falls after a decimal point, that zero is significant.
- When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
- When adding and subtracting, the answer is limited to the number of decimal places in the number with the least number of decimal places.
Hello!
The correct answer is 1. KCI.
I really hope this helped you out! c:
The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].
Formic acid is a weak acid. Thus, together with its conjugate base (formate) they form a buffer system. We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
![pH = pKa + log \frac{[formate]}{[formic\ acid]} \\\\5.00 = 3.75 + log \frac{[formate]}{[formic\ acid]}\\\\\frac{[formate]}{[formic\ acid]} = 17.8](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%20%5C%5C%5C%5C5.00%20%3D%203.75%20%2B%20log%20%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%5C%5C%5C%5C%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%20%20%3D%2017.8)
As expected, at a pH above the pKa, the concentration of formate is higher than that of the formic acid.
The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].
Learn more: brainly.com/question/22821585
