AlCl3 + Na NaCl + Al Did Na change oxidation number?

1. Define fission and fusion. Give an example of fission and fusion reflecting in real life application.

Alja [10]

Answer :

Nuclear fission : In nuclear reaction, the nucleus of a larger atom breaks into two or more smaller nuclei. In fission process, protons and neutrons are produced and larger amount of energy is released.

Example : In nuclear power plant, the energy released from the process of nuclear fission which is converted into electrical energy that is used in our homes and factories.

Nuclear fusion : In nuclear reaction, the nuclei of two or more smaller atoms combine together to form single larger molecule. In fusion process, the mass of the resulting nuclei is more as compared to the starting nuclei and large amount of energy is also released.

Example : This process occurs in the sun and stars. In this, the isotopes of Hydrogen, Tritium and Deuterium combine together to form a neutron and a helium atom under high pressure and temperature.

4 0

2 years ago

A mixture which contains more than one phase is called .

kirza4 [7]

This is a heterogeneous mixture.

6 0

3 years ago

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If the volume of 1.00 mole of gas is tripled, what will happen to the pressure, if the temperature remains constant

Julli [10]

Answer:

The new pressure becomes one third of the initial pressure.

Explanation:

The relation between pressure and volume at constant temperature is given by :

$P\propto \dfrac{1}{V}\\\\P=\dfrac{1}{V}$

Let new pressure and volume be P' and V' respectively.

V'=3V (given)

So,

$P'=\dfrac{1}{V'}\\\\=\dfrac{1}{3V}\\\\=\dfrac{1}{3}\times \dfrac{1}{V}\\\\P'=\dfrac{1}{3}\times P$

Hence, new pressure becomes one third of the initial pressure.

8 0

2 years ago

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

Answer:

For a: The empirical formula of the compound is $P_2O_5$

For b: The empirical formula of the compound is $KH_2PO_4$

Explanation:

For a:

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

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

Mass of P = 43.6 g

Mass of O = 56.4 g

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

Step 1: Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

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 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

For b:

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

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

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

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

Step 1: Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

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 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago

Calculate the amount of CaO that would be produced when 35.5g of CaCO3 decompose on heating (Ca= 40, O = 16, C = 12, H = 1

DanielleElmas [232]

amount of CaO that would be produced when 35.5g of CaCO3 decompose on heating (Ca= 40, O = 16, C = 12, H = 1

just put the values to have answer

4 0

3 years ago