A asystem at equilibrium stops
Answer: (Structure attached).
Explanation:
This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.
The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.
Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).
The reaction occurs in three steps:
- The attack on the electrophile forms the sigma complex.
- The loss of a proton regenerates an aromatic ring.
- The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).
Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.
<u>Answer:</u> The number of moles of strontium bicarbonate is 
<u>Explanation:</u>
Formula units are defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is 
We are given:
Number of formula units of 
As,
number of formula units are contained in 1 mole of a substance.
So,
number of formula units will be contained in =
of strontium bicarbonate.
Hence, the number of moles of strontium bicarbonate is 
Answer:
B.
Explanation:
electrons can be lost by one particle, and gained by another particle
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3