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xxMikexx [17]
4 years ago
14

Which of the following is a common industrial use for a mineral?

Chemistry
2 answers:
ELEN [110]4 years ago
7 0
The best and most correct answer among the choices provided by your question is the fourth choice.

A common use of minerals to industry is: <span>Using low-grade diamonds to cut materials .</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Leni [432]4 years ago
6 0

Answer: Option (d) is the correct answer.

Explanation:

It is known that a diamond is the hardest material according to Mohs scale. It also has highest thermal conductivity and that is why it is used in industrial applications.

It is used in the cutting and polishing of other tools or materials.  

Thus, we can conclude that out of the given options, using low-grade diamonds to cut materials is a common industrial use for a mineral.

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Answer: 2.52 x 10^(-10)

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A baseball is thrown a distance of 20 meters. What is its speed if it takes 0.5seconds to cover the distance
satela [25.4K]
The answer is that the baseball travels 40 miles per hour
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3 years ago
What is the mass of H2O produced when 10.0 grams of H2 reacts completely with 80.0 grams of O2?
Artyom0805 [142]
H₂:

M=2g/mol
m=10g

n = m/M = 10g/2g/mol = 5mol

O₂:

M = 32g/mol
m = 80g

n = m/M = 80g/32g/mol = 2,5mol

2H₂        +            O₂          ⇒          2H₂O
2mol            :       1mol        :          2mol
5mol            :       2,5mol     :          5mol
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B
8 0
3 years ago
For water ∆H°vap = 40.7 kJ/mol at 100.°C, its boiling point. Calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and
damaskus [11]

Answer : The value of change in entropy for vaporization of water is 1.09\times 10^2J/mol

Explanation :

Formula used :

\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}

where,

\Delta S^o = change in entropy  of vaporization = ?

\Delta H^o_{vap} = change in enthalpy of vaporization = 40.7 kJ/mol

T_b = boiling point temperature of water = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}

\Delta S^o=\frac{40.7kJ/mol}{373K}

\Delta S^o=1.09\times 10^2J/mol

Therefore, the value of change in entropy for vaporization of water is 1.09\times 10^2J/mol

5 0
3 years ago
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