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)
Answer:
0.0184
Explanation:
Let's consider the following reaction at equilibrium.
2 HI(g) ⇌ H₂(g) + I₂(g)
The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.
Kc = [H₂] × [I₂] / [HI]²
Kc = (4.78 × 10⁻⁴) × (4.78 × 10⁻⁴) / (3.52 × 10⁻³)²
Kc = 0.0184
The answer is C............
Answer: One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.
To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl.
<h3>HOPE THIS HELPS HAVE A AWESOME DAY❤️✨</h3>
Explanation:
Answer: place a 2 in front of NaNo3 on left side of equation while leaving the other blanks empty or you can place a 1 in those blanks
Explanation:
Step 1 count and write down the amount of each given element for both sides
Step 2 begin placing numbers (coefficients) to each side to balance
