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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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What is the mass of 5.00 moles of KaS04?
Varvara68 [4.7K]

Answer: D

Explanation:

I assume you meant \text{K}_{2}\text{SO}_{4}.

  • The atomic mass of potassium is 39.0983 g/mol.
  • The atomic mass of sulfur is 32.065 g/mol.
  • The atomic mass of oxygen is 15.9994 g/mol.

So, the formula mass of potassium sulfate is 2(39.0983)+32.065+4(15.9994)=174.2592 g/mol.

So, 5.00 moles have a mass of (5.00)(174.2592), which is about <u>870 g</u>

7 0
2 years ago
On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
Delicious77 [7]

Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

5 0
3 years ago
Read 2 more answers
Which of the following are properties of metalloids?
harkovskaia [24]

Answer:

all of these are properties of metalloids

3 0
2 years ago
Read 2 more answers
If 0.959J of heat is added to 0.88g of water, how much will its temperature increase
ss7ja [257]

Answer:

0.260 Celsius

Explanation:

q =c x m x (T2-T1)

c - specific heat of water 4.186 J/g.C

T2-T1 = q /(c x m) = 0.959 /(4.186 x 0.88) = 0.959/3.68 =0.260 C

7 0
2 years ago
The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.
avanturin [10]

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

6 0
3 years ago
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