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3 years ago

Why do we study isotopes?

1 answer:

3 0

Isotopes are atoms with different number of electrons
isotopes are very useful
for example isotope of uranium is used in atomic bombs and nuclear reactors radioactive isotopes are used to date ricks soils and archaeological objects.
isotopes have many uses that's why we study isotopes to know their uses

You might be interested in

How is the molecule of substance formed

rosijanka [135]

Answer:

When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.

Explanation:

8 0

3 years ago

1. The s orbitals are not symmetrical in shape. a. TRUE O. FALSE

ira [324]

The s orbitals are not symmetrical in shape is a FALSE statement.

An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.

The atomic orbitals in the atoms of elements differ in shape.

In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.

All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.

The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.

Read more:

brainly.com/question/5087295

4 0

2 years ago

A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87% H by mass. What is the empirical formula of this

Dmitry [639]

Answer: The empirical formula of the compound becomes $C_5H_{10}O_2$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Let the mass of the compound be 100 g

Given values:

% of C = 58.8%

% of H = 9.87%

% of O = [100 - 58.8 - 9.87] = 31.33%

Mass of C = 58.8 g

Mass of H = 9.87 g

Mass of O = 31.33 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

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

Step 1: Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol$

$\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol$

$\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol$

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

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles

$\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5$

$\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5$

$\text{Mole fraction of O}=\frac{1.96}{1.96}=1$

Converting the mole fraction into whole numbers by multiplying them with 2.

$\text{Mole fraction of C}=2.5\times 2=5$

$\text{Mole fraction of H}=5\times 2=10$

$\text{Mole fraction of O}=1\times 2=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 5 : 10 : 2

Hence, the empirical formula of the compound becomes $C_5H_{10}O_2$

8 0

3 years ago

2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons

Julli [10]

Answer :

The oxidation state of tin changes from (+2) to (+4).

The number electrons transferred in the reaction are, 10 electron.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The oxidation-reduction half reaction will be :

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $BrO_3^-\rightarrow Br_2$

First balance the main element in the reaction.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $2BrO_3^-\rightarrow Br_2$

Now balance oxygen atom on both side.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+H_2O\rightarrow Br_2$

Now balance hydrogen atom on both side.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+3H_2O\rightarrow Br_2+12OH^-$

Now balance the charge.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O+2e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The charges are not balance on both side of the reaction. We are multiplying oxidation reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $5SnO_2^{2-}+5OH^-\rightarrow 5SnO_3^{2-}+5H_2O+10e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The balanced chemical equation in basic medium will be,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The number electrons transferred in the reaction are, 10 electron.

7 0

3 years ago

A neutral atom of potassium (K) has an average mass of 39 amu and 19 electrons. How many neutrons does it have?

LenaWriter [7]

I believe the answer is 19

8 0

3 years ago

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