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bekas [8.4K]
3 years ago
7

Enter your answer in the provided box. On a certain winter day in Utah, the average atmospheric pressure is 718 torr. What is th

e molar density (in mol/L) of the air if the temperature is −29°C?
Chemistry
1 answer:
AnnZ [28]3 years ago
7 0

Answer:

The correct answer is 0.047 mol/L

Explanation:

The atmospheric air is a mixture of gases. We can assume an ideal behavior of the gas and use the ideal gas equation:

PV= nRT

where P is the pressure, V is the volume, n is the number of moles, R is a constant (0.082 L.atm/K.mol) and T is the temperature in K.

We have to first convert the pressure from Torr to atm:

760 Torr= 1 atm

⇒ 718 Torr x 1 atm/760 Torr = 0.945 atm

Then, we convert the temperature from ºC to K:

0ºC = 273 K

⇒ -29ºC+273= 244 K

Finally, we introduce the data in the equation and calculate de densitiy, which is the moles per liters of gas (n/V):

PV = nRT

n/V= P/RT

n/V = (0.945 atm)/(0.082 L.atm/K.mol x 244 K) = 0.047 mol/L

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Find the initial temperature of 45.0g of water if its final Temperature after submerging a metal with the mass of 8.5g is 22.0°C
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The initial temperature of the water that resulted in the final temperature of the water-metal mixture is 20.7 ⁰C.

<em>"Your question is not complete, it seems to be missing the following information;"</em>

the specific heat capacity of the metal is 0.45 J/g⁰C.

The given parameters;

  • <em>mass of water, </em>m_w<em> = 45 g</em>
  • <em>final temperature of the water, </em>t_w<em> = 22 ⁰C</em>
  • <em>mass of the metal, m = 8.5 g</em>
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The initial temperature of the water will be calculated by applying the principle of conservation of energy;

<em>heat gained by water = heat lost by metal</em>

Q_{water} = Q_{metal}

m_wc_w \Delta t_w = mc\Delta t

where;

c_w <em>is the specific heat capacity of the water = 4.184 J/g⁰C.</em>

<em />

<em>Substitute the given values;</em>

45 x 4.184 x (22 - t) = 8.5 x 0.45 x (85 - 22)

4142.16  - 188.28t = 240.98

188.28t  = 4142.16  - 240.98

188.28t  = 3901.18

t = \frac{3901.18}{188.28} \\\\t = 20.7 \ ^0 C

Thus, the initial temperature of the water that resulted in the final temperature of the water-metal mixture is 20.7 ⁰C.

Learn more here:brainly.com/question/15345295

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