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Andre45 [30]
3 years ago
13

When the sun shines directly on objects, white objects remain cooler than black objects. This difference in temperature is becau

se more of the light striking white objects is–
Chemistry
1 answer:
Dimas [21]3 years ago
6 0

Answer:

The amount of sun that is being reflected and how hot the sunlight being reflected is

Explanation:

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Consider the reaction below to answer the following questions (4) a. The nucleophile in the reaction is _______ b. The Lewis aci
KATRIN_1 [288]

The question is incomplete, the complete question is shown in the image attached to this answer.

Answer:

a) Br^-

b) FeCl3

c) slower

d) see the first attached image

Explanation:

Aromatic compounds undergo electrophilic substitution sections in the presence of the appropriate electrophile.

In the reaction above, the Br^- nucleophile attacks the Lewis acid FeCl3. Recall that the nitro group is meta directing hence the incoming Br^+ electrophile is directed towards the meta position as shown in the image attached.

Note that the nitro group deactivates the ring towards electrophilic substitution hence the reaction is slower with nitrobenzene than with unsubstituted benzene.

5 0
3 years ago
9)
Molodets [167]
The correct answer is D
7 0
3 years ago
Read 2 more answers
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

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.

5 0
3 years ago
What has fewer organisms then a domain but it also has more organisms then a phylum
kramer

Answer:

a species

Explanation:

6 0
3 years ago
Which best describes what forms in nuclear fission?
Effectus [21]

Answer: the joining of two nuclei. the splitting of a nucleus. the process by which the sun makes energy. the production of material to make atomic weapons.

Explanation:

6 0
4 years ago
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