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Dmitriy789 [7]
3 years ago
11

What dissolved species are present in a solution of naclo4? express your answers as ion separated by a comma?

Chemistry
2 answers:
Zina [86]3 years ago
7 0

<u>Answer:</u> The species that are present in a solution of sodium perchlorate are Na^+\text{ and }ClO_4^- ions.

<u>Explanation:</u>

We are given:

An ionic compound having chemical formula NaClO_4

Ionic compound is defined as the compound which is formed when electron gets transferred from one atom to another atom. These are usually formed when a metal reacts with a non-metal or a metal reacts with a polyatomic ion or a reaction between two polyatomic ions takes place.

This chemical compound is formed by the combination of sodium ions and perchlorate ions.

When sodium perchlorate is dissolved in water, it dissociates into its ions.

The chemical equation for the ionization of sodium perchlorate follows:

NaClO_4(aq.)\rightarrow Na^+(aq.)+ClO_4^-(aq.)

Hence, the species that are present in a solution of sodium perchlorate are Na^+\text{ and }ClO_4^- ions.

Afina-wow [57]3 years ago
4 0

The dissolved species that are  present  in solution of NaClO4 are Na+ ,   ClO4-

<h3>Explanation</h3>

when substance dissolve in water it dissociate  in ions. Some substance dissociate partially   when  they dissolve in water while others dissociate completely. NaClO4  is  a salt which dissociate completely   when it dissolve in water to give Na+ and CIO4-  ions

<h3 />
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List two general properties of molecular compounds
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Low melting points and boiling points. ...Low enthalpies of fusion and vaporization These properties are usually one or two orders of magnitude smaller than they are for ionic compounds.Soft or brittle solid forms. ...Poor electrical and thermal conductivity.
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Study the graph about oxygen content of Earth’s atmosphere. Oxygen Content of Earth apostrophe s Atmosphere line graph. X axis i
Anika [276]

Explanation:

from the graph study about oxygen content of Earth's atmosphere, we can understand that  

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The first cells on the earth are anaerobic microorganisms, as the CO2 level is too high they survive by using CO2.

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Starting around 2.7 billion years ago, photosynthesis by Cyanobacteria and later plants , pumped “OXYGEN” in to the atmosphere. This caused the decline of anaerobic bacteria and allows the diversification of animals as seen in “CAMBRIAN” around 500 millions year ago.  

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Modern human activities has raised both “CO2 and METHANE” level in the atmosphere to over leading to higher temperature and extinction of other species.

3 0
3 years ago
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The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has
Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate law expression is:

Rate=k[NO]^2[O_2]

Now we have to determine the number of molecules of NO\text{ and }O_2

In mixture 1 : There are 5 NO and 4 O_2 molecules.

In mixture 2 : There are 7 NO and 2 O_2 molecules.

In mixture 3 : There are 3 NO and 5 O_2 molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

Rate=k[NO]^2[O_2]

Rate=k(5)^2\times (4)

Rate=k(100)

The rate law expression for mixture 2 is:

Rate=k[NO]^2[O_2]

Rate=k(7)^2\times (2)

Rate=k(98)

The rate law expression for mixture 3 is:

Rate=k[NO]^2[O_2]

Rate=k(3)^2\times (5)

Rate=k(45)

Hence, the mixture 1 has the fastest initial rate.

4 0
3 years ago
The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

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Answer:

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