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Elodia [21]
3 years ago
15

A scuba diver at the surface of the sea takes a deep breath and fills her lungs with 6 liters of air. The pressure in her lungs

is 1 atm. She then dives down ten meters, without inhaling or exhaling, until the pressure of the air in her lungs is 2 atm. According to Boyle’s law, what happens to the air in her lungs as she dives down?
A. The temperature increases.
B. The volume increases.
C. The volume decreases.
D. The temperature decreases.
Chemistry
1 answer:
vazorg [7]3 years ago
7 0
<span>The answer is C. The volume decreases. A scuba diver at the surface of the sea takes a deep breath and fills her lungs with 6 liters of air. The pressure in her lungs is 1 atm. She then dives down ten meters, without inhaling or exhaling, until the pressure of the air in her lungs is 2 atm. According to Boyle’s law,  the air in her lungs decreases as she dives down.  </span>Boyle's Law<span> is the gas </span>law which states that in a closed space, pressure and volume are inversely related. As pressure increases, volume decreases, and vice versa.
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3 A soil has 70% sand, 20% silt
Kisachek [45]

Today, as part of the series of posts on soils, we are going to look at ‘soil texture’. Soil forms the basis for all life but it’s important to know about its mineral constitution as well as its biological profile.

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7 0
3 years ago
When adding or subtracting deelmals, how many digits should the answer contain?
Alexus [3.1K]

Answer:

Depends, but in most cases, 2.

It's best to use as many digits as possible to keep it accurate.

Explanation:

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This is something that depends in your situation.

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3 years ago
When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
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