Thomas poured 25 mL of 0.60 M HCl into a large bottle. He then added enough water so that the new (diluted) concentration was 0.
1 answer:
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Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq= =14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5=
14.5 x (0.15 M) x = (CH₃OH)
0.2818 M = (CH₃OH)
Answer:
------->
+
Explanation:
In equation 1, equating the mass number (A) on both sides.
A = 235 + 1 = A + 94 + 3*1
236 = A + 94 + 3
A = 236 - 94 = 3
A = 139
Equate the atomic numbers on both sides
92 + 0 = Z + 36 + 3*0
92 = Z + 36
Z = 92 - 36
Z = 56
In reaction 2, equating the mass number on both sides
235 + 1 = A + 143 + 3 *1
236 = A + 143 + 3
236 = Z + 146
Z = 90
Equatoing the atomic number of both sides
92 + 0 = Z + 54 + 3*0
92 = Z + 54
Z = 92 - 54
Z = 38.