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3 years ago

What does the atomic number ask

Chemistry

2 answers:

Alik [6]3 years ago

5 0

I'm not sure what it asks, but it answers

how many protons are in the nucleus
how many electrons are in the atom
where it sits in the periodic table
If the periodic table is divided in the modern way, it gives a good indicator of whether you are speaking about a metal, a non metal or neither.

timurjin [86]3 years ago

3 0

how many protons are in the nucleus

how many electrons are in the atom

where it sits in the periodic table

If the periodic table is divided in the modern way, it gives a good indicator of whether you are speaking about a metal, a non metal or neither.

This is what it answers, I hope that this helps

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Please help me l don’t understand please !!!!!!

garik1379 [7]

Answer: C

Explanation:

he never had evidence in the first place that was nearly enough.

7 0

2 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

PLEASE HURRY!!!

Bezzdna [24]

Answer:

Measurement of the amount of variation of the species in a given area.

Explanation:

Bio preffix means life. Thus, literally, biodiversity means how diverse is the life. This is, how many different organisms an ecosystem or biome has.

Thus, biodiverstiy, although more complex than just that, is a measure of the number of species that live in a region.

The biome with most species than any other, this is the biome with the greatest biodiversity, is the tropical rainforest biome, which are located near the equator. As you can imagine, a large number of different plants, animals, and microorganisms live in these forests, making them the regions with greatest biodiversity on the planet.

6 0

3 years ago

Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the energy of a single photon of this r

denis23 [38]

Answer :

(1) The frequency of photon is, $3\times 10^{10}Hz$

(2) The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = $1.0cm=0.01m$ (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

$\lambda$ = wavelength of photon

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$\nu=\frac{3\times 10^8m/s}{0.01m}$

$\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz$ $(1Hz=1s^{-1})$

The frequency of photon is, $3\times 10^{10}Hz$

(2) Now we have to calculate the energy of photon.

Formula used :

$E=h\times \nu$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})$

$E=1.988\times 10^{-23}J/photon$

The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) Now we have to calculate the energy in J/mol.

$E=1.988\times 10^{-23}J/photon$

$E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)$

$E=11.97J/mol$

The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

3 0

3 years ago

Read 2 more answers

For the 52nd electron must pair with one of the electrons in the 5p orbital.

Umnica [9.8K]

Answer:

This question is incomplete

Explanation:

This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as

[Kr] 4d¹⁰ 5s² 5p⁴ where [Kr] is the electronic configuration of krypton. Based on this, we can deduce that the 52nd electron will be in the first orbital of the P subshell (as attached in the picture). This is because when indicating the electrons in the subshell, one electron will be spread across each orbital and if any electron is still remaining, it will be added starting from to the first orbital of the subshell, however no two electrons in an orbital in a subshell can have the same spin and hence must face opposite direction based on pauli's exclusion principle (as seen in attached); thus for the 5p-orbital of elements with 52 or more electrons, when one electron each is represented in each box (3 boxes in total) in the 5p-orbital, the remaining electron is paired with the the first electron in the first box of the 5p-orbital

5 0

3 years ago

What does the atomic number ask​

What does the atomic number ask