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SVEN [57.7K]
3 years ago
7

Which statement accurately describes why air pressure decreases as altitude increases? As you increase altitude the temperature

difference between the layers of the atmosphere causes extreme winds and the air pressure in turn decreases. As you increase altitude, the air particles move around much more quickly because they have to travel further distances which decreases their pressure. As you increase altitude the temperature is increased and as you heat particles they spread further apart decreasing their pressure. As you increase altitude, Earth's gravity is not acting as greatly as when closer to Earth's surface so the weight of air molecules decrease, which also in turn decreases air pressure.
Chemistry
1 answer:
labwork [276]3 years ago
3 0

Answer:

The correct answer is " As you increase altitude, Earth's gravity is not acting as greatly as when closer to Earth's surface so the weight of air molecules decrease, which also in turn decreases air pressure."

Explanation:

As altitude increases air becomes rarefied. At higher altitudes there is less Earth's gravity acting on the air molecules than at areas close the earth surface. Air pressure is the weight of air acting per unit surface area. At higher altitudes as the weight of air decreases due to reduced gravitational pull the air pressure also decreases.

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Melissa's mother was making chocolate chip cookies, but told Melissa that she shouldn't eat the raw cookie dough. Why is it unsa
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the answer to the question is D
4 0
3 years ago
Make the following conversions. Show your calculations.
Natalija [7]

Answer:

1. 500 cm = 5 m

2. 1000 g = 1 kg

3. 455 L = 45, 500 cL

4. 0.865 m = 0.00865 m or 8.65 * 10^-3 n

5. 33.5 cm = 335 mm

6. 0.0198 m = 19800 micrometers

7. 57.65 cg = 5.765 * 10^8 nanograms

8. 1000 L = 1 kl

9. 99 degrees F = 37.222 degrees C

Explanation:

1.  One meter = 100 centimeters

500 centimeters/ 100 centimeters = 5 meters

2. One kilogram = 1000 grams.

1000 grams/1000 grams = 1 kilogram

3. One Liter = 100 centiliters

455 liters * 100 centiliters = 45, 500 centiliters

4. One meter = 100 centimeters

0.865 centimeters/100 centimeters = 0.00865 meters

5. One centimeter = 10 millimeters

33.5 centimeters * 10 millimeters = 335 millimeters

6. One meter = 1.0 * 10^6 micrometers

0.0198 meters * 1 * 10^6 micrometers = 19800 micrometers.

7. One centigram = 1.0 * 10^7 nanograms

57.65 centigrams * 1 * 10^7 nanograms = 5.765 * 10^8 nanograms.

8. One kiloliter = 1000 L

1000 L/ 1000 L = 1 kiloliter

9. The formula for finding Celsius is 5/9(f - 32)

So 5/9(99 - 32) = x

5/9(67) = x

x = 335/9 or 37.222 degrees Celsius.

5 0
3 years ago
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
Lion 7
Sidana [21]

Answer:

A.

Explanation:

You should NEVER eat or drink anything in a lab area. You never know what chemicals or gases are in the lab, and they can harm you.

Wearing a drawstring hoodie won't protect you from chemicals.

Don't wait to clean up chemicals, immediately get a teacher and clean it up (follow the teachers instructions). You never know what has spilled, and if it is harmful or not, or if there is a certain procedure to clean it up.

Don't change the equipment in the middle of an experiment. This can tamper with your results, and depending on what you are working with, this can be dangerous.

3 0
3 years ago
Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha
nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L

The moles of air are:

3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}

3 0
3 years ago
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