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Aleks04 [339]
2 years ago
12

Avanti works in a bookstore. She has four books and is going to place them in two stacks. The diagram above shows the books befo

re they touch. Use the information in the diagram to answer the question. How did the temperatures of the four books compare before they touched? How will the temperatures of the two top books compare after the books have been touching for a while, and why?

Chemistry
1 answer:
Feliz [49]2 years ago
3 0

Answer:

Well they didn't transfer any energy when they weren’t touching and it did t produce any energy if it didn’t move. Since they are on top of each other they are causing momentum on each other creating kinetic energy

Explanation:

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Calculate the final Celsius temperature when 634 L at 21 °C is compressed to 307 L.
Illusion [34]

Answer:

- 130.64°C.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>

<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>

5 0
3 years ago
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma
Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

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Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

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