Nitrogen in the limiting reactant x
Answer:
0.2425 M
Explanation:
Given mass = 9.36 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given that:- Volume = 965 mL = 0.965 L ( 1 mL = 0.001 L )
So,
Considering:
Answer:
Las moléculas de los reactivos tienen que chocar entre sí. Estos choques deben de producirse con energía suficiente de forma que se puedan romper y formar enlaces químicos. En el choque debe haber una orientación adecuada para que los enlaces que se tienen que romper y formar estén a una distancia y posición viable.
Answer is: B. C(s) + 2S(s) + 89.4 kJ → CS2(l).
Missing question:
A. C(s) + 2S(s) → CS2(l) + 89.4 kJ.
B. C(s) + 2S(s) + 89.4 kJ → CS2(l).
C. C(s) + 2S(s) + 89.4 kJ → CS2(l) + 89.4 kJ.
D. C(s) + 2S(s) → CS2(l).
Because enthalpy of
the system is greater that zero, this is endothermic reaction (<span>chemical reaction that
absorbs more energy than it releases)</span>, heat is included as a reactant.
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
