Answer:
Kc for this reaction is 0.43
Explanation:
This is the equilibrium:
N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)
And we have all the concentration at equilibrium:
N₂: 0.25M
H₂ : 1.3M
NO: 0.33M
H₂: 1.2M
They are ok, because they are in MOLARITY. (mol/L)
Let's make the expression for Kc
Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)
Kc = (0.33² . 1.2²) / (0.25 . 1.2²)
Kc = 0.4356
In two significant digits. 0.43
<span>30.0 ml of 0.15 m K2CrO4 solution will have more potassium ions.
Let's see the relative number of potassium ions for each solution. Since all the measurements are the same, the real difference is the K2CrO4 will only have 2 potassium ions per molecule while the K3PO4 solution will have 3 potassium ions per molecule.
K2CrO4 solution
30.0 * 0.15 * 2 = 9
K3PO4 solution
25.0 * 0.080 * 3 = 6
Since 9 is greater than 6, the K2CrO4 solution will have more potassium ions.</span>
Answer: On losing 6 moles of water, cobalt chloride forms unstable violet-coloured ions, before generating its stable blue-coloured anhydrous form.
Explanation:
The hydrated cobalt chloride loses its 6 water of crystallization, then dissociates into ions: cobalt ions and chlorine ions that appear violet, and quickly combined to form the stable anhydrous Cobalt chloride with blue colour.
Answer:
Explanation:
From the question we are told that:
Density of acetic acid 
Density of Water 
Generally the equation for Solution Density is mathematically given by
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
