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3 years ago

Elaborate on how to classify elements and pure compounds.

2 answers:

6 0

tthe corrrect answer is C)Elements and pure compounds are homogeneous materials because they have a uniform composition throughout.

tankabanditka [31]3 years ago

4 0

Answer: C) Elements and pure compounds are homogeneous materials because they have a uniform composition throughout.

Explanation: Element is a pure substance which is composed of atoms of similar elements. Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Elements are compounds form homogeneous materials as they have uniform composition throughout and the components are evenly distributed throughout the material.

Mixtures are heterogeneous materials as they do not have uniform composition and the components are not evenly distributed throughout the material.

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4 years ago

Which molecule is methylamine?

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An amine refers to a nitrogen-based compound. Since A is the only option with nitrogen, that would be the answer.

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3 years ago

Read 2 more answers

What is one of the possible products of a solution if cai2 is mixed with a solution of nano3?

11Alexandr11 [23.1K]

Answer:
The two products that will be formed are:
Ca(NO3)2
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4 years ago

How many elements had been discovered by 1860

UkoKoshka [18]

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3 years ago

Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is

Strike441 [17]

Answer:

For a: The molarity of commercial HCl solution is 12.39 M.

For b: The molality of commercial HCl solution is 16.79 m.

For c: The volume of commercial HCl solution needed is 2.42 L.

Explanation:

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

For a:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

For b:

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

For c:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

5 0

3 years ago