52 g of hydrogen H₂
Explanation:
I will assume that the problem is talking about the hydrogenation of carbon dioxide CO₂ not carbon monoxide CO. It is harder to reduce carbon dioxide than carbon monoxide and if you manage to reduce carbon dioxide you can reduce the carbon monoxide as well.
This reaction it will take place in the presence of catalyst at a specific temperature and pressure.
CO₂ + 4 H₂ → CH₄ + 2 H₂O
Now taking into the account the chemical reaction we devise the following reasoning:
if 1 mole of CO₂ react with 4 moles of H₂
then 6.5 moles of CO₂ react with X moles of H₂
X = (6.5 × 4) / 1 = 26 moles of H₂
number of moles = mass / molecular wight
mass = number of moles × molecular wight
mass of H₂ = 26 × 2 = 52 g
Learn more about:
hydrogenation reaction
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I am not entirely certain, but endothermic is a reaction in which more is kept than released, so likely not D. Sorry I couldn't be of more help, but you could probably just google endothermic reactions and get some of the information necessary. Good luck.
Answer:
I believe your answer would be, "A" A particle with two or more atoms bonded covalently.
Explanation:
The definition of a molecule;
A group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
Answer:
Qc <Kc and the system will evolve to the right.
Explanation:
The reaction constant Qc is calculated when a reaction that has not yet reached equilibrium and allows determining where the reaction will move to reach equilibrium. For the following reaction:
aA + bB ⇔ cC + dD
the constant Qc is calculated as:
![Qc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%2A%5BD%5D%5E%7Bd%7D%20%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Comparing Qc with Kc allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
In this case:
![Qc=\frac{[NaCl]*[H_{2}O] }{[HCl]*[NaOH]}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BNaCl%5D%2A%5BH_%7B2%7DO%5D%20%7D%7B%5BHCl%5D%2A%5BNaOH%5D%7D)
Being:
- [NaCl] = 1.75 M
- [H₂O] = 1 M
- [HCl] = 2.5 M
- [NaOH] = 3.5 M
Replacing:

Solving:
Qc=0.2
Being Kc = 0.5, then <u><em>Qc <Kc and the system will evolve to the right.</em></u>