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alexira [117]
3 years ago
9

Which characteristic is used to identify minerals?

Chemistry
2 answers:
koban [17]3 years ago
8 0

Answer:

b. location c. luster d. taste

Explanation:

  • The following physical and chemical properties for identifying the mineral are Color, Hardness, Luster, Taste, Odor, Etc. The properties help in identifying the place of origin the mineral has come from.  
  • In turn, location can be also used to identify a particular mineral or chemical origin. A mineral located in groundwater reserves has deposits of calcium and hence formed through the limestone rocks of the CaCo3 element.
  • Luster the reflectance properties of the rocks as some minerals don't have lust is called as "earthy," or "dull." And the color is an important property of the mineral and hence is a major tool in identifying the chemical and physical properties.
  • Other properties being Magnetism, Opaqueness and Specific Gravity.
snow_tiger [21]3 years ago
5 0
The ways to identify a mineral are LUSTER, density, hardness, and cleavage. 
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The grams of solute are required.

The mass of solute is 3.5 g

c =  Molarity = 0.1 M

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n = Number of moles

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Molarity is given by

c=\dfrac{n}{V}\\\Rightarrow n=cV\\\Rightarrow n=0.1\times 0.1\\\Rightarrow n=0.01\ \text{moles}

Molar mass is given by

M=\dfrac{m}{n}\\\Rightarrow m=Mn\\\Rightarrow m=350\times 0.01\\\Rightarrow m=3.5\ \text{g}

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A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
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Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

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Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

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Hence, work for path A is -2818.68 J.

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Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

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Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

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