In 3.8 moles of li, there are about <span>0.144071459 grams. I assume that you are talking about lithium.</span>
The answer is D. Okay l hope this helps
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
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Answer:
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
Step-by-step explanation:
The unbalanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + KNO₃
Notice that the complex groups like NO₃ and CrO₄ stay the same on each side of the equation.
One way to simplify the balancing is to replace them with a single letter.
(a) For example, let <em>X = NO₃</em> and <em>Y =CrO₄</em>. Then, the equation becomes
PbX₂ + K₂Y ⟶ PbY + KX
(b) You need 2X on the right, so put a 2 in front of KX.
PbX₂ + K₂Y ⟶ PbY + 2KX
(c) Everything is balanced. Now, replace X and Y with their original meanings. The balanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
