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Flauer [41]
1 year ago
9

You work for a catering company who just got an order in for 5 giant fruit salads. If you make the giant fruit salads according

to the equation above, will you be able to fill the order if you have 10 pineapples, 20 cantaloupes, and 75 bunches of grapes?A.) No, the cantaloupes will be limiting.B.) No, the grapes will be limiting.C.) No, the pineapples will be limiting.D.) Yes, you have enough of everything.

Chemistry
1 answer:
borishaifa [10]1 year ago
5 0

We need to find out how many fruits we need to make 5 giang fruit salads. For this, we use the rule of 3:

1 giant fruit salad --- 2 pineapples

5 giant fruit salad --- x pineapples

x = 10 pineapples

1 giant fruit salad --- 3 cantaloupes

5 giant fruit salad --- x cantaloupes

x = 15 cantaloupes

1 giant fruit salad --- 16 bunches of grapes

5 giant fruit salad --- x bunches of grapes

x = 80 bunches of grapes

As we can see, we have exactly the right amount of pineapples, 5 more cantaloupes, and 5 fewer bunches of grapes.

It means that cantaloupes are in excess and the grapes are limiting.

Answer: No, the grapes will be limiting.

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Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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