Important (question on k12 test)

Chemistry

2 answers:

monitta3 years ago

4 0

Answer:

A new Dana system of classification contains 78 different classes of minerals based on composition and then further classified by type and group. To be considered a mineral, a substance must be an inorganic, naturally formed solid, with a specific chemical formula and a fixed internal structure. To test whether something is a mineral, there are several identification tests to which the substance is subjected, including its resistance to scratching, its density in comparison to water, its color, the degree of light it reflects, the color of the powdered mineral, its breakage pattern and its crystalline form.

Explanation:

Luden [163]3 years ago

4 0

Answer: chemical makeup and crystal lattice structure

Explanation:

The minerals are the inorganic substances that are extracted beneath the earth crust. These minerals are associated with the gangue particles that are required to be removed by processing.

The minerals can be classified on the basis of the chemical and physical properties. The chemical properties define the chemical composition of the mineral. The physical properties define how minerals react with the external environment.

Thus chemical makeup and crystal lattice structure can be a useful approach to classify the minerals as the minerals may differ chemically as well as in structural makeup.

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A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance

sasho [114]

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$ , where K is the Coulomb´s constant ( $9x10^{9} N\frac{m^{2} }{C^{2} }$ ), $q_{1}$ = Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$ .

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$