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vodka [1.7K]
3 years ago
11

What is the total volume occupied by 132 g of CO2 at STP

Chemistry
2 answers:
ad-work [718]3 years ago
7 0
<span>132g x (22.4L/44 grams) = 67.2L
The answer = 67.2 L 
Hope it helps!
Puffin</span>
Lilit [14]3 years ago
4 0
132 grams x (1 mol / 44 grams) = 3 moles 
<span>3 moles X (22.4 L/ 1 mol) = 67.2 </span><span>L</span>
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3 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
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Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

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         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

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According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

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3 years ago
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<span>People with protanopia are unable to sense any ‘red’ light, people with deuteranopia do not sense ‘green’ light and people with tritanopia cannot sense ‘blue’ light. If a person perceives the color green, then the yellow sensitive nerves must work somewhat effectively since green is a combination of yellow and blue. Red-sensitive nerves are most likely not responding properly for this person. The answer is C.</span>
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Answer:

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Explanation:

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Answer:

D because i did this last week and got it right.

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