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Aleksandr [31]
2 years ago
10

J. A gas has a volume of 12 L at 200°C with 6.0 atm.

Chemistry
2 answers:
myrzilka [38]2 years ago
8 0

Answer:

0.4 \:

Explanation:

I only know this answer hope it will help you if it's wrong I am sorry:(

kai6417

White raven [17]2 years ago
4 0

Gay Lussac's Law

\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

Convert:

200°C = 200 + 273 = 473 K

50°C = 50 + 273 = 323 K

Input the value:

\tt \dfrac{6}{473}=\dfrac{P_2}{323}\\\\P_2=4.097~atm

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1) What is air pressure? (4 Points)
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<span>3)    </span><span>How does air pressure change?  <span>(Answer: Millibar values used in meteorology range from about 100 to 1050. At sea level, standard air pressure in millibars is 1013.2. Weather maps showing the pressure at the surface are drawn using millibars. ... This change in pressure is caused by changes in air density, and air density is related to temperature.)</span></span>

<span>4)    </span><span>Why is cooler, drier air related to High Pressure? <span>(Answer: This is due to density differences between the two air masses. Since stronger high-pressure systems contain cooler or drier air, the air mass is denser and flows towards areas that are warm or moist, which are in the vicinity of low pressure areas in advance of their associated cold fronts.)</span></span>

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kolbaska11 [484]

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4 0
3 years ago
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

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8 0
3 years ago
How much pressure does an elephant with a mass of 2200 Kg and a total footprint area of 4500 cm2 exert on the ground?
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Answer:

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2200 kg is 2200*9.8=21560 N, and 4500 cm^2=4500/10000=0.45 m^2.

So the total pressure exerted on the ground (!!) is 21560/0.45= 47911.1 Pa.

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3 years ago
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