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3 years ago

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

Chemistry

1 answer:

Alja [10]3 years ago

4 0

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.

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3 years ago

Dimethyl sulfoxide [(ch3)2so], also called dmso, is an important solvent that penetrates the skin, enabling it to be used as a t

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The molecular formula of dimethyl sulfoxide is $(CH_{3})_{2}SO$ . Molar mass of dimethyl sulfoxide is 78.13 g/mol. Calculate number of moles as follows:

$n=\frac{m}{M}=\frac{7.14\times 10^{3} g}{78.13 g/mol}=91.38 mol$

From the molecular formula, 1 mole of dimethyl sulfoxide contains 2 moles of Carbon, 6 moles of Hydrogen, 1 mole of Sulfur and 1 mole of oxygen.

Thus, 91.38 moles of dimethyl sulfoxide will have:

Carbon :

$n_{C}=2\times 91.38 moles=182.77 moles$

Hydrogen:

$n_{H}=6\times 91.38 moles=548.28 moles$

Sulfur:

$n_{S}=1\times 91.38 moles=91.38 moles$

Oxygen:

$n_{O}=1\times 91.38 moles=91.38 moles$

Since, 1 mole of an element equals to $6.023\times 10^{23}$ atoms thus, number of atoms can be calculated as:

Carbon:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$182.77 moles\rightarrow 182.77\times 6.023\times 10^{23} atoms=1.10\times 10^{26} atoms$

Hydrogen:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$548.28 moles\rightarrow 548.28\times 6.023\times 10^{23} atoms=3.30\times 10^{26} atoms$

Sulfur:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms$

Oxygen:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms$

Therefore, number of C, S, H and O atoms are $1.10\times 10^{26}, 5.50\times 10^{25}, 3.30\times 10^{26} and 5.50\times 10^{25} atoms$ respectively.

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4 years ago

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2 years ago

What is the solubility (in g/L) of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10–11

PilotLPTM [1.2K]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$CaF_2\rightleftharpoons Ca^{2+}+2F^-$