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3 years ago

Atoms of oxygen-16, oxygen-17 and oxygen-18 are isotopes. Compare and contrast these three isotopes.

Chemistry

1 answer:

maria [59]3 years ago

6 0

Answer:

There are three isotopes of the element oxygen (O): Oxygen 16, 17, and 18. Each isotope of oxygen contains 8 protons, but differs in the number of neutrons. ... Therefore, oxygen 16 has 8 protons and 8 neutrons, oxygen 17 has 8 protons and 9 neutrons, and oxygen 18 has 8 protons and 10 neutrons.

Chemical elements are found in different versions, called isotopes. Isotopes are elements that contain the same amount of protons, but differ in the number of neutrons in their nuclei. For example, there are three isotopes of the element oxygen (O): Oxygen 16, 17, and 18. Each isotope of oxygen contains 8 protons, but differs in the number of neutrons. An isotope number is a shorthand representation of its mass. Because protons and neutrons are roughly equal in mass, an isotope’s number is equal to the sum of its protons and neutrons. Therefore, oxygen 16 has 8 protons and 8 neutrons, oxygen 17 has 8 protons and 9 neutrons, and oxygen 18 has 8 protons and 10 neutrons.

There are two main types of isotopes that geoscientists use to interpret the ancient Earth: stable and unstable isotopes.

Explanation:

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Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$