Answer:
all behaved the same
Explanation:
I can't explain I'm just trying
Answer:
a) the reverse reaction is favoured
b) the forward reaction is favoured
c) the forward reaction is favoured
Explanation:
The equation ought to have been correctly written as;
3A + 2B --------> C + 2D. ∆H =20 kJ
Actually, we can see that the reaction is endothermic since ∆H= positive.
We know that when pressure is decreased, the reaction tends towards the side with higher total volumes. There are five volumes(moles) of reactants and three volumes(moles) of products. A decrease in pressure will favour the reverse reaction.
Being an endothermic reaction, increase in temperature is known to favour the forward reaction. Similarly, removing D will drive the equilibrium forward thereby favouring the forward reaction.
All the fossil fuels...coal, oil, gas...are examples of no return enable energy resources.
Answer:
First step would be convert to moles
Final Answer: 37.8 g of NaCl
Explanation:
The reaction is:
2Na + Cl₂ → 2NaCI
We convert the mass of each reactant to moles:
18 g . 1mol /23g = 0.783 moles of Na
23g . 1mol / 70.9g = 0.324 moles of chlorine
We use the mole ratio to determine the limiting reactant:
Ratio is 2:1. 2 moles of Na react to 1 mol of chlorine
Then, 0.783 moles of Na, may react to (0.783 . 1)/2 = 0.391 moles.
Excellent!. We need 0.391 moles of Cl₂ and we only have 0.324 moles available. That's why the Cl₂ is our limiting reactant.
We use the mole ratio again, with the product side. (1:2)
1 mol of Cl₂ can produce 2 moles of NaCl
Then, our 0.324 moles of gas, may produce (0.324 . 2)/1 = 0.648 moles
Finally, we convert the moles to grams:
0.648 mol . 58.45g/mol =
This is an incomplete question, here is a complete question.
A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸
Answer : The concentration of 
ions at equilibrium is, 
Explanation : Given,
Moles of 
= 0.130 mol
Volume of solution = 1 L
Concentration of = Concentration of = 0.130 M
Concentration of 
= 1.20 M
The equilibrium reaction will be:
![Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq%29%2B6NH_3%28aq%29%5Crightarrow%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D)
Initial conc. 0.130 1.20 0
At eqm. x [1.20-6(0.130)] 0.130
= 0.42
The expression for equilibrium constant is:
![K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%5BNi%28NH_3%29_6%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D%5D%5BNH_3%5D%5E6%7D)
Now put all the given values in this expression, we get:
Thus, the concentration of ions at equilibrium is,
