The arrows B, C, and E represent the process of cooling.
Solid state: In the solid state, the molecules are arranged in a regular and fixed pattern. The molecules in a solid are closely packed that is the solid particles can not move.
Liquid state: In the liquid state, the molecules are present in an irregular pattern. The molecules are closely packed but particles can move.
Gaseous state: In the gaseous state, the molecules are present in an irregular manner. The molecules of gases are not closely packed and can move freely.
Therefore, the solid changes to liquid on heating, and the liquid changes to solid after cooling.
Liquid changes to a gaseous state on heating while gases changes to liquid on cooling.
Solid changes to gas on heating and gases change to solid after cooling.
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Answer:
Kc for this equilibrium is 2.30*10⁻⁶
Explanation:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.
Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is defined as:
![Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%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)
In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.
In this case, being:
2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)
the equilibrium constant Kc is:
![Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BN_%7B2%7D%20%5D%2A%5BH_%7B2%7D%20%5D%5E%7B3%7D%20%20%7D%7B%5BNH_%7B3%7D%20%5D%5E%7B2%7D%20%7D)
Being:
- [N₂]= 0.0551 M
- [H₂]= 0.0183 M
- [NH₃]= 0.383 M
and replacing:
you get:
Kc= 2.30*10⁻⁶
<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>
Answer: 20 mg Te-99 remains after 12 hours.
Explanation: N(t) = N(0)*(1/2)^(t/t1/2)
N(t) = (80 mg)*(0.5)^(12/6)
N(t) = 20 mg remains after 12 hours
Answer:
C). Half-reactions with SRP values greater than zero are spontaneous.
Explanation:
SRPs or Standard Reduction Potentials are characterized as the ability of a probable distinction among the anode and cathode of a usual/standard cell. It aims to examine the capacity of chemicals to reduce themselves.
The third statement asserts <u>a true claim regarding the SRPs(Standard Reduction Potentials) that the 'half-reactions which take place with the SRP possesses the values higher than zero and they are unconstrained</u>.' The other statements are incorrect as they either show the estimation of SRPs more than 0 or display them as being restricted. Thus, <u>option C</u> is the correct answer.
1,1-hexene > 2,3,3-dimethyl-2-butene > 3-methyl-3-hexene > cis-3-hexene
The Saytzeff rule states that an alkene becomes more stable the more highly substituted it is. Since 2,3-dimethyl-2-butene is the most substituted alkene among the ones listed, it will thus be the most stable.
Ozone treatment of 2,3-dimethyl but-2-ene results in ozoinde, which with further reduction yields propanone and water. Trans-2-butene undergoes ozonolysis, producing the main ozonide. Acetaldehyde, syn- and anti-acetaldehyde oxide, and this main ozonide are the products of its breakdown.
Let's not forget that the alkenes' pi bonds are least stabilized by alkyl groups, making terminal alkenes the least stable of the group. This means that among the alkenes listed, 1-hexene is the least stable.
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