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mixas84 [53]
3 years ago
7

What does homogeneous mixture of elements

Chemistry
1 answer:
photoshop1234 [79]3 years ago
3 0

a homogeneous mixture is a mixture that appears to be one thing. for example air would be one.

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Where does lava in volcanoes come from?
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Magma can push through holes or cracks in the crust of the volcano, causing a volcanic eruption. Which is when magma flows or erupts onto earth's surface, which is what you call lava (whenever it goes onto the earths surface).
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Will the solubility of CaSO4 in 0.10 Molar CaF2 be smaller than, larger than, or the same as in pure water?
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7. Jasmine wants to prevent getting burned by accidently touching a hot metal pipe, What would be the worst
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A) Glass

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3 years ago
A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21 c and the pressure is lowered to 0.80 atm,
Nuetrik [128]

Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

\frac{P*V}{T}=k

Having an initial state 1 and a final state 2 it is possible to say that:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

  • P1= 1 atm
  • V1= 1.2 L
  • T1= 273 °K
  • P2= 0.80 atm
  • V2= ?
  • T2= 21°C= 294 °K

Replacing:

\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}

Solving:

V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}

V2= 1.62 L

<u><em>The new volume is 1.62 L</em></u>

<u><em></em></u>

8 0
2 years ago
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