Answer:
5.The limiting reactant is completely used up in the reaction
Explanation:
The limiting reactant is completely used up in the reaction is the correct answer because a limiting reactant is a reactant in chemical reaction that is completely consumed or used up in the chemical reaction. Limiting reactant when it is completely used up limits the amount of products that will be formed. The reaction will be halted or will stop when the limiting reactant is totally used up.
Answer:
speed and how slidey things are
Explanation:
Answer:-
Neon is a noble gas. Neon with an electronic configuration of 2,8 has 8 electrons in it's outermost shell or valence shell.
According to the Octet theory if an element has eight electrons in it's valence shell then it is stable and does not undergo reaction. Thus Neon does not need to react and can exist freely in nature.
Sodium with an electronic configuration of 2,8,1 has 1 electron in it's valence shell. As per octet rule, it is not stable and it must lose that 1 electron to become stable.
In order to lose that 1 electron sodium must react with other substances. Hence sodium cannot exist freely in nature.
Answer:
A) 
.
B) 
.
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)
B) For the calculation of Kc, we rate the equilibrium expression:
![Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent 
, we have:
![[NH_3]=0.6M=2*x](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.6M%3D2%2Ax)
Next, the concentrations of nitrogen and hydrogen at equilibrium are:
![[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M](https://tex.z-dn.net/?f=%5BN_2%5D%3D%5Cfrac%7B1.5mol%7D%7B1.5L%7D-x%3D1M-0.3M%3D0.7M)
![[H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B4mol%7D%7B1.5L%7D-3%2Ax%3D2.67M-0.9M%3D1.77M)
Therefore, the equilibrium constant is:
C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.
Answer:
the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
Explanation:
Given that;
The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K
Initial volume of the container, V1 = 537 mL
Initial vapor pressure, P1 = 68.0 mmHg
Final volume of the container, V2 = 338 mL
Let us say that the final vapor pressure = P2
From Boyle's law,
P2V2 = P1V1
P2 * 338 = 68.0 * 537
338
P2 = 36516
P2 = 36516 / 338
P2 = 108.03 mmHg
Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
