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Alchen [17]
3 years ago
10

As a science major, you are assigned to the "Atomic Dorm" at college. The first floor has one suite with one bedroom that has a

bunk bed for two students. The second floor has two suites, one like the one on the first floor, and the other with three bedrooms, each with a bunk bed for two students. The third floor has three suites, two like the ones on the second floor, and a third suite that has five bedrooms, each with a bunk bed for two students. Entering students choose room and bunk on a first-come, first-serve basis by criteria in the following order of importance:
(1) They want to be on the lowest available floor.
(2) They want to be in the smallest available dorm room.
(3) They want to be in a lower bunk if available.

(a) Which bunk does the 1st student choose?
(b) How many students are in top bunks when the 17th student chooses?
(c) Which bunk does the 21st student choose?
(d) How many students are in bottom bunks when the 25th student chooses?
Chemistry
1 answer:
lidiya [134]3 years ago
8 0

Explanation:

A.

The first student will be on the lower bunk on the first floor because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available.

B.

7 students are in the TOP bunks because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available. Therefore, all the rooms up till the third floor (Remember, third floor has 3 suites), so the first floor is filled - 1 person on the top bunk, 2 floor is filled- 4 persons and the third floor; the first suite is filled - 1 person and the second suite is a little partially filled- 1 person.

C.

Following the criteria 1, 2 and 3, the 21st student occupies the third suite on the third floor because all the floors (1 and 2) are occupied so the third suite on the third floor is still vacant.

D.

From the criteria there are therefore 10 persons at the TOP bunk. All the rooms up till the third floor are filled, so the first floor is filled - 1 person on the top bunk, second floor is filled (2 suites) - 4 persons and the third floor; the first suite and second suite is filled - 4 persons; the thirs suite has 6 persons present so 1 person is at the top bunk.

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Volume:

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mass : 213 g

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
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