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kaheart [24]
3 years ago
11

Imagine a friend is planning a rock-climbing trip. Write a note in the box explaining how gas exchange is affected at the top of

a mountain, where air pressure is lower and there is less oxygen than at lower altitudes.
Chemistry
1 answer:
wariber [46]3 years ago
6 0

Answer:

Air is less dense on a mountaintop than at sea level.

Air pressure is lower at low altitudes.

As you climb a mountain, air pressure increases.

More force pushes on the air at the bottom of an air column.

As you descend a mountain, air molecules are closer together.

Explanation:

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A chemical reaction produces formaldehyde, with a chemical formula of CH2O. Carbon is in Group 4A, oxygen is in Group 6A, and hy
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This problem is providing a comparison on the atoms forming formaldehyde, CH₂O, in terms of the groups they are in, C in group 4A, O in group 6A and H in group 1A, and it is asking for a description about the bonds once they are bonded in a formal Lewis dot structure.

In such a way, the attached file shows the Lewis dot structure, in which we can find two bonds between the two hydrogen atoms and the carbon one, and also, a double bond between carbon and oxygen.

The aforementioned is explained by bearing to mind the concept of octet, which states that elements in groups 4A, 5A, 6A and 7A are able to complete 8 valence electrons after bonding.

Thus, since carbon is in group 4A we infer it needs 4 more electrons to complete the octet, and that is achieved due to the 4 bonds it is having with the two hydrogen atoms and double bond with the oxygen atom.

In addition, since oxygen is in group 6A, we infer it needs two more electrons to complete its octet, which is attained via the double bond it has with the carbon atom. For hydrogen, however, it can complete one bond at maximum, because it is in group 1A, which means it can complete two valence electrons after bonding.

Learn more:

  • brainly.com/question/10535983
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4 0
3 years ago
In the image of the billiard table below, a cue ball is about to be struck and pushed toward the other balls
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Explanation:

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3 years ago
A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is
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<span>Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
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