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3 years ago

Describe the particles that make up a pure substance

Chemistry

1 answer:

Anni [7]3 years ago

8 0

Answer:

Each element consists of indivisible, minute particles called atoms.

All atoms of a given element are identical.

Atoms of different elements have different masses.

Explanation:

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How many cubic miles are 8.48E+08 gallons of water? The density of water at ambient conditions is 1.000 g/mL.

Inessa05 [86]

The volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .

<h3>Weight of one gallon of water</h3>

The weight of 1 gal of water is given as 3785 g

Mass of 8.48 x 10⁸ gal = 3785 x 8.48 x 10⁸ = 3.2 x 10¹² g

<h3>Volume of the water in cubic meters</h3>

Volume = mass/density

Volume = 3.2 x 10¹² g/1 gmL

Volume = 3.2 x 10¹² mL x 10⁻⁶ m³/mL = 3.2 x 10⁶ m³

Thus, the volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .

Learn more about volume here: brainly.com/question/1972490

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7 0

2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

Atoms with loosely held valence electrons have 1. high ionization energy and high electron affinity. 2. high ionization energy a

EastWind [94]

Answer:

Explanation:

Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.

The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low

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3 years ago

In what type of reaction will an acid and a base react with each other

Arada [10]

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acidcombines with the OH(-) anion of the base to form water.

6 0

3 years ago

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Which statement best explains the net transfer of energy that is about to

ArbitrLikvidat [17]

Answer:

Explanation:

B I got it right

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3 years ago

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