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Leona [35]
3 years ago
12

What are the 9 properties of minerals and explain each property?​

Chemistry
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

hardness-Hardness is the ability of a material to resist deformation

luster-the way light interacts with the surface of a crystal

color-the quality of an object or substance with respect to light reflected by the object

streak- the color of the powder produced when it is dragged across an un-weathered surface.

specif gravity-the density of that object divided by the density of water cleavage -the tendency of a mineral to break along smooth planes

fracture-any separation in a geologic formation

tenacity-refers to a mineral's toughness or resistance to breaking or being deformed.

lol sorry that took awhile

You might be interested in
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
4 years ago
The reaction, 2C4H10 (g) + 13O2 (g) à 8CO2 (g) + 5H2O (g), is the combustion of butane. What occurs as the reaction proceeds?
Solnce55 [7]
The reaction, 2 C4H10 (g) + 13 O2 (g) = 8 CO2 (g) + 5 H2O (g), is the combustion of butane.   A combustion reaction involves the reaction of a hydrocarbon with oxygen producing carbon dioxide and water. This reaction is exothermic which means it releases energy in the form of heat. Therefore, as the reaction proceeds,a heat energy is being given off by the reaction. This happens because the total kinetic energy of the reactants is greater than the total kinetic energy of the products. So, the excess energy should be given off somewhere which in this case is released as heat.
7 0
3 years ago
Help, please!
professor190 [17]

Hi so from what I can see the pizza as a distant or and you just have to convert the grams of glucose into moles. Most teachers ask for this format

7 0
3 years ago
A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of
inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 \frac{g}{c m^{3}}

To find:

Mass of the copper cylinder

<u>Step by Step by explanation:</u>

Solution:

According to the formula, Mass can be calculated as

\rho=\frac{m}{v} and from this

m=\rho \times v

Where, m=mass of the cylinder

\rho =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

v=\pi r^{2} h

Where \pi=3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, v=3.14 \times 12^{2} \times 44

v=3.14 \times 144 \times 44

v=19895.04 \mathrm{cm}^{3}

And we know that, m=\rho \times v

Substitute the known values in the above equation we get

m=8.90 \times 19895.04  

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

4 0
4 years ago
Choose whether each statement is a scientific hypothesis, theory, or law. The pressure a gas exerts on a container increases as
Naya [18.7K]

Answer:

1) theory

2) law

3) hypothesis

Explanation:

edge 2021

7 0
3 years ago
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