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horsena [70]
2 years ago
11

Figure 2 shows what happens to the electrons in the outer shells when a sodium atom reacts with a fluorine atom.

Chemistry
1 answer:
QveST [7]2 years ago
4 0

Sodium reacts with fluorine to make a covalent bond between them.  A covalent bond is formed when there is a mutual sharing of electrons between the two adjacent atoms.

Here, Sodium is metal and fluorine is non-metal. So, the bond between them may be an ionic bond, but as the Sodium has one valence electron and fluorine is highly electronegative. Sodium wants to lose electrons and fluorine wants to gain electrons to form a chemical bond. Therefore, the sodium donates its valence electron to fluorine, forms a cation, and fluorine accepts the electron and forms an anion and they distribute the difference in the charge between them.

So, we can say that when sodium and fluorine react to form sodium fluoride with a covalent bond.

Learn more about covalent bonds here:

brainly.com/question/12732708

#SPJ10

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A gas with a volume of 250 mL at a temperature of 293K is heated to 324K. What is the new volume of the gas?
Natalija [7]

Answer:

The new volume of the gas is 276.45 mL.

Explanation:

Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.

Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T} =k

Analyzing an initial state 1 and a final state 2, it is satisfied:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1= 250 mL
  • T1= 293 K
  • V2= ?
  • T2= 324 K

Replacing:

\frac{250 mL}{293 K} =\frac{V2}{324 K}

Solving:

V2=324 K*\frac{250 mL}{293 K}

V2= 276.45 mL

<em><u>The new volume of the gas is 276.45 mL.</u></em>

4 0
3 years ago
Ni-cad (nickel–cadmium) batteries have a slightly lower cell potential than the common alkaline value of 1.5 v normally used in
Lorico [155]
The half-reaction are:

Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.

NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.

Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.

ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>


6 0
2 years ago
When oxygen reacts with hydrogen, it has the capacity to release 29 kilojoules of energy. Inside a fuel cell, oxygen reacts with
Vanyuwa [196]

Answer:

79

Explanation:

7 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
NEED HELP!!!
Vilka [71]

Answer:

1.atomic number

2.electron

3.element

4.atom

5.neutron

6.nucleus

7.proton

Explanation:

please like and Mark as brainliest

6 0
2 years ago
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