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andrey2020 [161]
2 years ago
9

A piece of iron (C=0.449 J/g°C) and a piece of gold (C=0.128 J/g°C) have identical masses. If the iron has an initial temperatur

e of 498 K and the gold has an initial temperature of 298 K, which of the following statements is TRUE of the outcome when the two metals are placed in contact with one another? Assume no heat is lost to the surroundings.
a. Since the two metals have the same mass, the final temperature of the two metals will be 398 K, exactly half way in between the two initial temperatures. b. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K. c. Since the two metals have the same mass, the thermal energy contained in the iron and gold after reaching thermal equilibrium will be the same. d. Since the two metals have the same mass, but the specific heat capacity of gold is much smaller than that of iron, the final temperature of the two metals will be closer to 298 K than to 498 K. e. None of the above are true.
Chemistry
1 answer:
spin [16.1K]2 years ago
4 0

Answer:

The correct answer is B. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K

Explanation:

Iron is hotter and gold is colder, therefore, according to laws of thermodynamics, iron will lose heat to gold until they are at the same temperature.

The specific heat capacity of iron(0.449) is over three times that of gold(0.128). Since masses are equal, this means that each time iron's temperature drops by one degree, the energy released it releases makes gold's temperature increase by more than 3 degrees. So gold's temperature will be climbing much faster than iron's is falling. Meaning they will meet closer to the initial temperature of iron than that of gold

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A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O
Alexxandr [17]
C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O  produces 24 + 4 + 32 g C2H4O2    (theoretical)
 46 g produces 60g 

60 g C2H4O2 is produced from 46g C2H6O
1g      .     .................................46/60 g 
 700g     .................................    (46/60) * 700  Theoretically

But as the yield is only 7.5% 

the required amount is    ((46/60) * 700 ) / 0.075 =  7155.56 g 

=  7.156 kg to nearest gram.  Answer




8 0
2 years ago
A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

8 0
2 years ago
Describe how to prepare the solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the
padilas [110]

Answer:

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

6 0
2 years ago
The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
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4 0
2 years ago
Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1
erastovalidia [21]

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

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Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, <u>in this case</u> / Total mass) . 100

<u>We were asked for the acetonitrile</u> → (667.7 g / 824 g) . 100 = 81%

3 0
2 years ago
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