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3 years ago

What type of molecule is shown below

Chemistry

1 answer:

Kobotan [32]3 years ago

4 0

Answer is: C alkane.

Name of this molecule is n-hexane.

Hexane is alkane (acyclic saturated hydrocarbon, carbon-carbon bonds are single) of six carbon atoms.

Carbon atoms in hexane have sp3 hybridization.

In sp3 hybridization hybridize one s-orbital and three p-orbitals of carbon atom.

Alkene has one double bond and alkyne has one triple bond.

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eimsori [14]

Distance and period of time

6 0

3 years ago

What volume of 0.194 MNa3PO4 solution is necessary to completely react with 85.5 mL of 0.109 MCuCl2 ?

ira [324]

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

4 0

2 years ago

Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the phy

barxatty [35]

Answer:

1. $S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

2. $Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

3.The electron affinity of $Mg^{2+}$ is zero.

4. $O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

Explanation:

<u>Electron affinity:</u>

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of $S^{-}$ is as follows.

$S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

<u>Ionization energy</u>:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

$Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

The electronic configuration of Mg: $1s^{2}2s^{2}2p^{6}3s^{2}$

By the removal of two electrons from a magnesium element we get $Mg^{2+}$ ion.

$Mg^{2+}$ has inert gas configuration i.e, $1s^{2}2s^{2}2p^{6}$

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of $Mg^{2+}$ is zero.

The ionization energy of $O^{2-}$ is follows.

$O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

3 0

3 years ago

What are the stages of the life cycle of the sun

Jet001 [13]

The Sun is currently a main sequence star and will remain so for another 4-5 billion years. It will then expand and cool to become a red giant, after which it will shrink and heat up again to become a white dwarf. The white dwarf star will run out of nuclear fuel and slowly cool down over many billions of years.

5 0

3 years ago

Read 2 more answers

2) Calculate the percent composition of each element in Mgso,

iogann1982 [59]

Answer:

$\% Mg=20.2\%\\\\\% S=26.6\%\\\\\% O=53.2\%$

$\% Ag=93.1\%\\\\\% O=6.9\%$

Explanation:

Hello!

2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:

$\% Mg=\frac{24.305}{120.36 } *100\%=20.2\%\\\\\% S=\frac{32.06}{120.36 } *100\%=26.6\%\\\\\% O=\frac{64.00}{120.36 } *100\%=53.2\%$

3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:

$\% Ag=\frac{215.74}{231.74} *100\%=93.1\%\\\\\% O=\frac{16.00}{231.74} *100\%=6.9\%$

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7 0

3 years ago