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Butoxors [25]
3 years ago
15

What is the smallest particle which prossesses the properties of a compound called?​

Chemistry
1 answer:
never [62]3 years ago
4 0

Answer:

Molecule

Explanation:

Compounds are chemical substances comprising of two or more elements. The smallest particle contained in a compound is the MOLECULE. Molecules are combinations of atoms (same or different) of an element held together by a chemical bond.

Since a compound is a molecule with two or more elements combined, the chemistry of the molecules contained in a compound determines its chemical properties. This means that the molecules of a compound are responsible for the interactions of a compound in a chemical reaction. Examples of molecules are H2O, N2, O3 etc.

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An isotope with a mass number of 193 has 116 neutrons. What is the atomic number of this
Zinaida [17]

Answer:

Atomic number of this  isotope = 77

Explanation:

Given that,

Mass number = 193

No of neutrons = 116

We need to find the atomic no of this isotope.

We know that,

Atomic mass = No of protons + No. of neutrons

Also, atomic no = no of protons

So,

Atomic mass = atomic no + No. of neutrons

⇒ Atomic no = Atomic mass - no of neutrons

Atomic no = 193 - 116

Atomic no = 77

Hence, 77 is the atomic no of the isotope.

6 0
3 years ago
What is the maximum amount of kcl that can dissolve in 200g of water?
Likurg_2 [28]
The solubility of potassium chloride in at room temperature is approximately 34 g per 100 g of water. Therefore, the maximum amount that could be dissolved would be 34/100 ( 200) = 68 g of KCl. When more than this amount is added, excess potassium would not dissolve forming crystals in the solution.
8 0
3 years ago
Gas is heated from 480. K to 750. K and the pressure is kept constant, what final volume would result if the original volume was
attashe74 [19]
Charles law gives the relationship between volume and temperature of gas.
It states that at constant pressure volume is directly proportional to temperature
Therefore
V/ T = k
Where V - volume T - temperature in kelvin and k - constant
V1/T1 = V2/T2
Parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
Substituting the values in the equation
267 L/ 480 K = V / 750 K
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8 0
3 years ago
Minor partial melting in the asthenosphere causes liquid magma to form. If the molten rock is able to move at all, it will leave
Bad White [126]
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6 0
3 years ago
When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

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=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

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

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

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

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> 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 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

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

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
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