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tensa zangetsu [6.8K]
2 years ago
10

Which of the following statements is true?

Chemistry
1 answer:
Fittoniya [83]2 years ago
8 0

Answer:

Es la tercera por qué cuando el material varía va cambiando de estado

Explanation:

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think about putting a glass of water on the counter, in the freezer, and on a hot stove how does the temperature of the water af
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The kinetic energy and the physical state of water depend strongly on the temperature;

  • Firstly, The kinetic energy of water on a hot stove is higher than that on the counter in the freezer; that the kinetic energy is directly proportional to the temperature according to the relation: K = \frac{3RT}{2NA} ; where R is the universal gas constant, T is the temperature and NA is Avogadro number.

As the temperature increases, the speed of colliding molecules increases and the kinetic energy increases.

  • Secondly, The physical state of water depends on the temperature; water has three states (gas, liquid and solid) depends on the temperature.
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Read 2 more answers
Calculate the kinetic energy of a 8.8 x 10^13 kg comet moving at a speed of 33.0 km/s. Round your answer to 2 significant digits
g100num [7]

Answer:

KE=4.79\times 10^{22}\ J

Explanation:

Given that

Mass ,m= 8.8 x 10¹³

Speed ,v= 33 km/s

We know that 1 km= 1000 m

v= 33 x 10³ m/s

As we know that kinetic energy given as

KE=\dfrac{1}{2}mv^2

Now by putting the values in the above equation

KE=\dfrac{1}{2}\times 8.8\times 10^{13}\times 33000^2

KE=4.7916\times 10^{22}\ J

Therefore we can say that the kinetic energy will be

KE=4.79\times 10^{22}\ J

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son4ous [18]
<span>According to Mendeleyev-Klapeyron’s equation pV = nRT, where p = 160 atm V = 12 R -constant 0.0821 & T = 298 in Kelvin Using given data, we can determine the amount of Helium gas: n = pV/RT = (160â™12)/(0,0821â™298) = 78,48 (mol) For atmospheric pressure (1 atm) and the same amount we can calculate the volume of tank, using previous equation: V = nRT/p = (78,48â™0,0821â™298)/1 = 1920 (liters) V = 1920 liters Thus Answer is 1920 liters</span>
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