Answer: 5. Positive
6. Negative
9. Oxygen carbon hydrogen nitrogen
Explanation:
<span>The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. </span>
Answer:
Reagents: 1) 
2) 
, 
Mechanism: Hydroboration
Explanation:
In this case, we have a <u>hydration of alkene</u>s reaction. But, in this example, we have an <u>anti-Markovnikov reaction</u>. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: <u>"hydroboration"</u>.
The <u>first step</u> of this reaction is the addition of borane () to the double bond. Then in the <u>second step</u>, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the <u>third step</u>, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond "
". In <u>step number 4</u> we have the migration of the C-B bond to oxygen. Then in <u>step number 5</u>, we have the attack of on the 
to produce an alkoxide. Finally, the water molecule produce in step 2 will <u>protonate</u> the molecule to produce the alcohol.
See figure 1
I hope it helps!
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Answer:
In both nuclear and chemical reactions, two physical quantities are seen to be conserved and unchanging: the number of particles and the total charge. A constant number of particles in nuclear reactions does not imply that mass is conserved.
Explanation:
