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3 years ago

The gold foil experiment led to the discovery of which subatomic particle?

Chemistry

1 answer:

tatuchka [14]3 years ago

3 0

Answer:

The gold foil experiment led to the discovery of nucleus in an atom

Explanation:

The structure of atom was first given by the plum pudding model of J.J. Thomson before the experiment of Ernest Rutherford. The plum pudding model explained an atom as a positive charge body which contains small negatively charged particles which are called electrons. He also described that the negative charge in atom is balanced with the equal amount of positive charge to maintain the neutrality of atom. But there were some faults in this model of Thomson. He did not give the complete structure of atom which was then given by Rutherford in his Gold Foil Experiment in 1898 which was published in 1911.

He discovered the concept of nucleus in atom. His research is based on the experiment with alpha particles. Alpha particles are helium atom particles. He did bombardment of positive alpha particles on thin foil of gold approx. 8.6 x 10 -6 centimetres thick and took the observations on the screen of zinc sulphide which was behind the gold foil. He observed the deflection of these bombarded alpha particles on the photographic film

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Plz help chemistry test

Nataly [62]

Answer:

plants take carbon dioxide out of the atmosphere and use the energy from sunlight to combine the carbon dioxide and water to form sugar and oxygen

4 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

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

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

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 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is $CH_2F$

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 = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

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

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

The formula for a compound of Li+ ions and Br- ions is written LiBr. Why can’t it be written Li2Br? Why isn’t it written BrLi?

maksim [4K]

Lithium has charge of +1 and bromide has charge of - 1. So they combine to form the compound lithium bromide which is expressed as LiBr.

<h3>Explanation:</h3>

Lithium is an alkali metal placed in group 1 or periodic table. It has a valency of 1 which is achieved as lithium loses an electron to achieve a charge of +1.

Bromine is a halogen which is placed in group 17 of periodic table. It has a valency of 1 which is achieved as bromine looses an election to achieve a charge of - 1.

Lithium is the cation and bromide is the anion. So lithium is written in front and bromine following the cation. And as both of their valencies are 1, so they form the compound LiBr.

3 0

3 years ago

Consider the following reactions. (Note: (s) = solid, () = liquid, and (g) = gas.)

White raven [17]

Answer:

Both are endothermic reactions.

Explanation:

Chemical equation:

1/2H₂(g) + 1/2I₂(g) → HI(g) + 6.2 kcal/mol

Chemical equation:

21.0 kcal/mol + C(s) + 2S(s) → CS₂

Both reaction are endothermic because heat is added in both of reactions.

Endothermic reactions:

The type of reactions in which energy is absorbed are called endothermic reactions.

In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.

For example:

C + H₂O → CO + H₂

ΔH = +131 kj/mol

it can be written as,

C + H₂O + 131 kj/mol → CO + H₂

Exothermic reaction:

The type of reactions in which energy is released are called exothermic reactions.

In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.

For example:

Chemical equation:

C + O₂ → CO₂

ΔH = -393 Kj/mol

it can be written as,

C + O₂ → CO₂ + 393 Kj/mol

6 0

4 years ago

Which block of elements do the halogens belong to? a. p block c. d block b. s block?

kykrilka [37]

A. p block

from 3a-8a are in the p block including halogen

4 0

3 years ago