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3 years ago

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Which of the following is consistent with a reaction that proceeds spontaneously in the reverse direction (assume all variables

are in terms of the forward direction only)?
A. ΔG° = 0, Q = K
B. ΔG > 0, Q > K
C. ΔG° > 0, Q = K
D. ΔG < 0, Q > K
E. ΔG > 0, Q < K

1 answer:

elena-s [515]3 years ago

8 0

Answer:

B. ΔG > 0, Q > K

Explanation:

For reverse reaction to take place Q should be greater than K

Since reverse reaction should be spontaneous, forward reaction should be non spontaneous.

So, for forward reaction, ΔG > 0

Hence, the correct answer is B

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A theater director reinterprets a play by replacing its classical score with

cricket20 [7]

Answer:

The play will be more appealing to a younger audience.

Explanation:

A younger audience will more likely appreciate current pop hits rather than classical score.

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2 years ago

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Mature elephants consume an average of 600 pounds of food per day. If an elephant

MAXImum [283]

Answer:

$\large \boxed{\text{4 600 000 kg or 4600 Mg}}$

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

$\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}$

3. Convert days to pounds of feed

$\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}$

4. Convert pounds to kilograms and megagrams

$\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}$

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

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3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

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3 years ago

When concentrated nitric acid mixed with some concentrated sulfuric acid reacts with benzene at moderate temperature forms------

FrozenT [24]

I will help just send me it

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2 years ago

Formed when an amine is combined with a carboxyl group

alekssr [168]

Hello.

The answer is C.Amine

When an amine is combined (reacted) with a carboxyl group, an AMIDE + water is formed, and if you carry on heating under a vacuum, an imidazoline is formed.

Have a nice day

7 0

3 years ago