Answer:
Q >> Kc
We have more products than reactans. To reach the equilibrium, the balance will shift to the left.
Explanation:
Step 1: Data given
Temperature = 450.0 K
Kc = 4.62
When Kc > Q, we have more reactants than products. To reach the equilibrium, the balance will shift to the right
When Kc < Q, we have more products than reactans. To reach the equilibrium, the balance will shift to the left.
When Kc = Q,the equiation isatequilibrium
[SO3] = 0.254 M
[O2] = 0.00855 M
[SO2] = 0.500 M
Step 2: The balanced equation
2SO2(g) + O2(g) ⇄ 2SO3(g)
Step 3: Calculate the Q
Q = [SO3]² / [O2][SO2]²
Q = 0.254²/ (0.500 * 0.00855²)
Q = 1765
Q >> Kc
We have more products than reactans. To reach the equilibrium, the balance will shift to the left.
Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = 
moles = 0.00600 moles
Neutralization reaction (back titration): 
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = 
moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.
Answer:
0.971 grams
Explanation:
Given:
Temperature = 3.0° C = 3 + 273 = 276 K
Volume, V = 5.0 L
Pressure, P = 0.100 atm
Now, from the relation
PV = nRT
where,
n is the number of moles,
R is the ideal gas constant = 0.082057 L atm/mol.K
thus,
0.1 × 5 = n × 0.082057 × 276
or
n = 0.022 moles
Also,
Molar mass of the Dinitrogen monoxide gas (N₂O)
= 2 × Molar mass of nitrogen + 1 × Molar mass of oxygen
= 2 × 14 + 16 = 44 grams/mol
Therefore, Mass of 0.022 moles of N₂O = 0.022 × 44 = 0.971 grams
Answer:
D.) They often form hydroxide ions.
Explanation:
They generate hydroxide ions in water. they are soapy to touch and bitter in taste. they conduct electricity.
(all bases have hydroxide ions)
