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castortr0y [4]
2 years ago
4

A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. Af

ter 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers.
Chemistry
1 answer:
aev [14]2 years ago
6 0

Answer:

The answer to your question is Second student

Explanation:

Let's pretend the data are:

500-ml bottle                        2-L bottle

T1 = 25°C                               T1 = 25°C

T2 = 4°C                                T2 = 4°C

Cp = 4.18 J/g°C

Density = 1 g/ml

then

mass 1 = 500 g                     mass 2 = 2000 g

Gradient of temperatures

ΔT = 4 - 25 = -21°C               ΔT = 4 - 25 = -21°C

Heat formula

Q = mCpΔT

Process

Calcule heat gained or lost in both situations

Q1 = (500)(4.18)((-21)            Q2 = (2000)(4.18)(-21)

Q1 = -43890 J                      Q2 = -175560 J

Conclusion: The second bottle lost more heat.

The first student was wrong because the heat lost depend on the mass quantity not the the initial or final temperature.

The second student was right, heat depends on the quantity of matter.

The third student was wrong because heat lost do not depend on the speed it is lost or gained.

The fourth student is wrong because we can stablish temperature as constant, and then only quantity of matter changes.

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How many grams of NH3 are needed to provide the same number of molecules as in 0.35 g of SF6 ?
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Answer:

<u />

  • <u>There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.</u>

<u />

Explanation:

Using the molar mass of the chemical formula SF₆ you can find the number of moles of molecules in 0.35 g of such substance. Then, using the molar mass of NH₃, you can find mass in grams corresponding to the same number of molecules.

<u>1. Find the molar mass of SF₆:</u>

Atom   atomic mass          number of atoms   total mass in 1 mole

S           32.065 g/mol                     1                                       32.065 g

F            18.998 g/mol                     6                 6 × 18.998 = 113.988 g

                                                                               =====================

                                                          molar mass of SF₆ =    146.053 g/mol

<u>2. Find the number of moles in 0.35 g of SF₆:</u>

  • number of moles = mass in grams / molar mass
  • number of moles = 0.35 g / 146.053 g / mol = 0.0024 mol

<u>3. Find the molar mass of NH₃:</u>

Atom   atomic mass          number of atoms   total mass 1 mole                

N           14.007 g/mol                     1                                       14.007 g

H             1.008 g/mol                    3               3 × 1.008 g = 113.988 g

                                                                               =====================

                                                         molar mass of NH₃ =    17.031 g/mol

<u />

<u>4. Find the mass in 0.0024 mol of NH₃:</u>

  • mass in grams = number of moles × molar mass

  • mass = 0.0024 mol × 17.031 g/mol ≈ 0.041 grams

<u>5. Conclusion: </u>

There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.

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2 years ago
Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a __________
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Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a ____________ mixture. A) heterogeneous
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A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine
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Answer:

The final volume of the sample of gas V_{2} = 0.000151 m^{3}

Explanation:

Initial volume V_{1} = 200 ml = 0.0002 m^{3}

Initial temperature T_{1} = 296 K

Initial pressure P_{1} = 101.3 K pa

Final temperature T_{2} = 336 K

Final pressure P_{2} =  K pa

Relation between P , V & T is given by

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

Put all the values in the above equation we get

101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )

V_{2} = 0.000151 m^{3}

This is the final volume of the sample of gas.

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2 years ago
The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

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