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Mrac [35]
3 years ago
9

In a hospital, oxygen is administered to patients at 3.0 atm in a hyperbaric oxygen chamber. Oxygen gas, measuring 600.0 L, is c

ompressed in a cylinder at 160.0 atm. What volume of oxygen can a cylinder supply at the given pressure?
Chemistry
1 answer:
atroni [7]3 years ago
5 0

Answer: ifkkk

Explanation:

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Which element is classified as a noble gas?
Fed [463]

Answer:

Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

Explanation:

I found my answer on google

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6 0
3 years ago
This number tells you the number of each type of atom in a compound.
KonstantinChe [14]

Answer:

avogadro's constant

Explanation:

this is the fixed number of the atoms in the molecule of an element

avogadro's law states that equal volumes of gases<em> </em><em>at</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em> </em><em>contain</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>

<em>that</em><em> </em><em>is</em><em> </em><em>all</em><em> </em><em>gases</em><em> </em><em>with</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em>will</em><em> </em><em>always</em><em> </em><em>have</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>

7 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
Please help ASAP! Will mark brainiest if correct
Scilla [17]

Answer: 1 is A

2 is C

3 is D

Hope this help

7 0
3 years ago
Name the following molecule in the image
Vesna [10]

\\ \sf{:}\longrightarrow 2,4-diethyl-2-methylpentane

4 0
2 years ago
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