First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹
The answer is C
Answer:
IV
Explanation:
The complete question is shown in the image attached.
Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.
Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.
Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.
Answer:
The answer is SiO2
Explanation:
Silocon dioxide is written without a 1 after the silocon and with a 2 after the oxygen.
Magnesium (Mg)
The reason for this is the reactivity of the listed metals. Gold and silver are extremely unreactive metals. It is because of this unreactive nature that they remain in good condition for long periods of time, and are preferred in jewelry. Copper, although more reactive than gold and silver, is still not reactive enough to react with HCl.
The only metal that will react is magnesium.
Answer:
5.7 moles of O2
Explanation:
We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:
2KClO3 —> 2KCl + 3O2
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.
Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.
Thus, 5.7 moles of O2 were obtained from the reaction.
