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Maru [420]
2 years ago
7

7. A movable piston is allowed to cool from 392°F to 104°F. If the initial volume is 105 mL,

Chemistry
1 answer:
-BARSIC- [3]2 years ago
5 0

Answer:

27.9 mL

Explanation:

To find the new volume, you need to use the Charles' Law equation:

V₁ / T₁ = V₂ / T₂

In this equation, "V₁" and "T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature.

V₁ = 105 mL                     V₂ = ? mL

T₁ = 392 °F                       T₂ = 104 °F

V₁ / T₁ = V₂ / T₂                                            <----- Charles' Law

105 mL / 392 °F = V₂ / 104 °F                     <----- Insert values

0.26785 = V₂ / 104 °F                                 <----- Simplify left side

27.9 = V₂                                                     <----- Multiply both sides by 104

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Vesna [10]

Answer:

9000 BC

Explanation:

Although various copper tools and decorative items dating back as early as 9000 BCE have been discovered, archaeological evidence suggests that it was the early Mesopotamians who, around 5000 to 6000 years ago, were the first to fully harness the ability to extract and work with copper.

3 0
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
artcher [175]

Explanation:

The given data is as follows.

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

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

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

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

          R = 8.314 J/K mol

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.

4 0
3 years ago
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Nimfa-mama [501]
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7 0
3 years ago
Read 2 more answers
In a synthesis reaction you start with 1.7L of Hydrogen how many liters of water will be produced?
jolli1 [7]

15.3 litres of water will be produced if we take 1.7 litres of Hydrogen

Explanation:

Let's take a look over synthesis reaction;

H_{2}+ O_{2}<u>                         </u>H_{2}O<u />

<u>Balancing the chemical reaction;</u>

2H_{2} +O_{2}<u>                        </u>2H_{2}O<u />

Thus, 2 moles of hydrogen molecules are required to form 2 moles of water molecules.

<u>Equating the molarity;</u>

<u />\frac{1.7*1}{2*2} = \frac{x*1}{2*18}

           (Since, the molecular mass of hyd and water is 2 and 18 respectively)

x=\frac{1.7*2*18}{2*2}

x= 15.3 litres.

Thus,15.3 L of water will be produced if we take 1.7 litres of Hydrogen in a synthesis reaction.

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