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iogann1982 [59]
3 years ago
11

Locate the element xenon (Xe) on the periodic table and click on the square. What class of element is xenon found in?

Physics
2 answers:
musickatia [10]3 years ago
4 0
Noble gas, that is the correct answer on e2020!
vaieri [72.5K]3 years ago
3 0

Answer : The element 'Xenon' belongs to the class of noble gas.

Explanation :

The element 'Xenon' with symbol 'Xe' belongs to group 18. The atomic number of xenon is, 54. It is a colorless, odorless noble gas which is found in the earth's atmosphere.

The electronic configuration of xenon (Xe) will be,

1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^6

Hence, the element 'Xenon' belongs to the class of noble gas.

The element xenon are shown by the red box in the periodic table.

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If you walk at an avaerage speed of 5 km/h for 30 min how far will you walk
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2.5 km/30min

Explanation:

Divide 5 by 2 because 30  min is half of 1 hour. when you do that you get 2.5

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On average, each person in the industrialized world is responsible for the emission of 10,000 kg of carbon dioxide () every year
frosja888 [35]

Answer:

17.2 meters

Explanation:

Carbon dioxide (CO2) molecular mass is 44g/mol, then the number of moles of 10000kg carbon dioxide will be:

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The volume of gas in STP is 22.4 L/mol, then the volume of the carbon dioxide will be:

227,272 moles* 22.4L/mol= 5,090,909 L

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7 0
3 years ago
A mass of 0.560 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic mo
Slav-nsk [51]

Answer:

(a) A = 0.700m

(b) k = 80.6N/m

(c) x = -0.699m

(d) x = -0.350m

(e) t = 0.168s

Explanation:

Given the equation of motion for the spring

X = 0.700cos(12.0t), m = 0.56kg

(a) A = amplitude = 0.700m

(b) The angular velocity ω = 12rad/s

ω = √(k/m)

ω² = k/m

k = m×ω² = 0.56×12² = 80.6N/m

Spring constant k = 80.6N/m

(c) T = 2π/ω = 2π/12

T = 0.524s

At t = T/2 = 0.524/2 = 0.262s

So x = 0.700cos(12×0.262) = –0.699m

(d) At t = 2/3×T = 2×0.524/3 = 0. 349s

x = 0.700cos(12×0.349) = –0.350m

(e) to find t at x = -0.300m

–0.300 = 0.700cos(12t)

–0.300/0.700 = cos(12t)

cos(12t) = –0.429

12t = cos-¹(-0.429)

12t = 2.01

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