Methods of Heat Transfer

Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate

Anvisha [2.4K]

Answer:

$-3272$ kJ/mol

Explanation:

Given and known facts

Mass of Benzene $= 0.187$ grams

Mass of water $= 250$ grams

Standard heat capacity of water $= 4.18$ J/g∙°C

Change in temperature ΔT $= 7.48$ °C

Heat

$=250 * 4.18 * 7.48\\=7816.6 \\=7.82$

Heat released by benzine is - 7.82 kJ

Now, we know that

$0.187$ grams of benzene release $= -7.82$ kJ heat

So, $1$ g benzine releases

$\frac{ -7.82 }{0.187}\\= -41.8$

kJ/g

$0.187 * \frac{1}{78.108}=0.00239$ mol C6H6

Heat released

$= \frac{-7.82}{ 0.00239}$

$=-3272$ kJ/mol

4 0

3 years ago

What type of chemist studies how materials conduct electricity?

e-lub [12.9K]

Explanation:

The study of interaction between the chemical substances is known as chemistry and the person who does this study is called a chemical scientist. Everything around us is made up of matter and the interaction between this matters for one or different substances.

Conduits lead electrical flow in all respects effectively as a result of their free electrons. Separators restrict electrical flow and make poor transmitters. Some basic channels are copper, aluminium, gold, and silver. Some regular encasing are glass, air, plastic, elastic, and wood.

3 0

3 years ago

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In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz,

Vika [28.1K]

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

$\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}$

$\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}$

$Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz$

Therefore, the signal is at 11759 Hz from the TMS.

6 0

4 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) THIS IS THE RIGHT ANSWER (the more negative= the more released)

7 0

3 years ago

Why is it necessary to use a new salt bridge for each salt bridge for each electrochemical cell?

Irina-Kira [14]

Answer:

salt bridge balances the charge when electrons move from one half cell to another half cell.

Explanation:

Explanation: A salt bridge balances the charge when electrons move from one half cell to another half cell. During this process the salt bridge uses its electrolyte solution which further helps in balancing charges in both the half cells. ... Therefore, for each electrochemical cell a new salt bridge is used.

6 0

3 years ago

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