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2 years ago

Finish the attachment pleaseee it's due I will give extra points and brainly too

Chemistry

1 answer:

arlik [135]2 years ago

4 0

Answer:

the first orbit contains a maximum of 2 electron

The second orbit contains a maximum of 8

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How many atoms are in 8.66 moles of barium? Round your answer to 2 decimal places.

yarga [219]

Answer:

52.15 × 10²³ atoms

Explanation:

Given data:

Number of moles = 8.66 mol

Number of atoms = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ atoms

8.66 mol × 6.022 × 10²³ atoms / 1mol

52.15 × 10²³ atoms

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3 years ago

What is an substance

lora16 [44]

a particular kind of matter with uniform properties.

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3 years ago

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What are the reasons for existence of large number of organic compounds in nature.Explain

Law Incorporation [45]

Answer:

because of catenation of carbon.

Explanation:

Catenation is the binding of an element to its self through covalent bonds to form chain or ring molecules. carbon is able to form continuous links with other carbon atoms which is the reason for the existence of a large number of organic compounds.

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3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

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3 years ago

A solid-solid solution is called a(n)

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3 years ago

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