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OlgaM077 [116]
3 years ago
10

1. Distinguish between a mixture and a pure substance. ​

Chemistry
2 answers:
Marina CMI [18]3 years ago
4 0

Answer:

1)Mixture is the combination of substance

2)There are two types of mixture such as heterogeneous mixture and homogeneous mixture

3)foreg=Salt solution ,sedimentation

Pure substance

1)Pure substances are also called elements

2)They are made up of small atoms

3)for eg= oxygen,Venadium ,Iron,Oxygen

Korvikt [17]3 years ago
3 0

Mixture - A mixture is a substance which consist of two or more elements or compounds not chemically combined together.

  • For Example : Air is a mixture of gases like oxygen,nitrogen,argon,carbon dioxide etc.

Pure Substance - A pure substance is a substance which has only one type of atoms or molecules thereby having a constant composition and structure.

  • For Example : Hydrogen gas , Gold metal , Sugar , Baking soda, Ammonia , Diamond and etc.
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1) Write the chemical equation.

CH_4+2O_2\rightarrow CO_2+2H_2O

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: <em>unknown</em>.

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<em>3.1- Set the equation.</em>

PV=nRT

<em>3.2- Plug in the known values and solve for n (moles).</em>

(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=n=0.24\text{ }mol\text{ }CH_4

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The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: <em>unknown</em>.

<em>5.1- Set the equation.</em>

PV=nRT

<em>5.2- Plug in the known values and solve for V (liters).</em>

(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}V=3.98\text{ }L

3.98 L of O2<em> is required to react with 2.0 L CH4.</em>

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