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gizmo_the_mogwai [7]
4 years ago
8

How many moles of gas are contained in a rigid, sealed, 5 L container, at 1 atm of pressure and 22 oc?

Chemistry
1 answer:
Ugo [173]4 years ago
7 0
The ideal gas law is: PV=nRT
Pressure
Volume
n= moles
R= gas constant
Temperature in Kelvin
(Degrees in celsius +273)
n= PV/RT
(1.00atm)(5.00L)/(.08026)(295K)= .207mol of gas
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oee [108]
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8 0
3 years ago
Which statement explains why low temperature and high pressure are required to liquefy chlorine gas?
Liula [17]
The correct answer is option 2. Chlorine gas molecules are held together by strong covalent bonds. This type of bond requires an ample amount of energy in order to break it and turn gas into liquid. That is why, high pressure and low temperature is needed.
5 0
3 years ago
As
Mars2501 [29]

Explanation:

  1. area of rectangle=l×b
  • 15=5×b
  • b=3cm

2.area of rectangle=l×b

  • 28=l×4
  • l=7cm

3. perimeter of square=4×L

  • 24=4×L
  • L=6cm

4. perimeter of rectangle=2(L+b)

  • 14=2(4+b)
  • 14=8+2b
  • 2b=6
  • b=3cm

hope it helps.

4 0
3 years ago
Read 2 more answers
Can someone help please?
attashe74 [19]
<h3>Take the weighted average of the individual isotopes.</h3><h3 /><h3>Explanation:</h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>69.2</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>65</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>30.8</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>So, the weighted average is </h3><h3>62.93</h3><h3>×</h3><h3>69.2</h3><h3>%</h3><h3> </h3><h3>+</h3><h3> </h3><h3>64.93</h3><h3>×</h3><h3>30.8</h3><h3>%</h3><h3> </h3><h3>=</h3><h3> </h3><h3>63.55</h3><h3> </h3><h3>amu</h3><h3> .</h3><h3 /><h3>If we look at the Periodic Table, copper metal (a mixture of isotopes but </h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> and </h3><h3>65</h3><h3>C</h3><h3>u</h3><h3> predominate) has an approximate atomic mass of </h3><h3>63.55</h3><h3> </h3><h3>g</h3><h3>⋅</h3><h3>m</h3><h3>o</h3><h3>l</h3><h3>−</h3><h3>1</h3><h3> , so we know we are right.</h3>

4 0
3 years ago
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Lera25 [3.4K]

Answer:

maybe the 3rd stance if your not sure try to search in googel

5 0
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