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EastWind [94]
3 years ago
14

Can someone help me with my chemistry homework please???

Chemistry
1 answer:
taurus [48]3 years ago
4 0
<h2><u>Answers:</u></h2>

<u>1.) Lithium and Sulfide: </u>

  • Formula: \bold{Li_{2}S}
  • Ion Charges: \bold{Li~1+,~Li~1+,~S~2-}
<h3 />

<u>2.) Lithium and Chlorine:</u>

  • Formula: \bold{2LiCl}
  • Ion Charges: \bold{Li~1+, Li~1+,Cl~1-,Cl~1-}
<h3 />

<u>3.) Lithium and Oxygen: </u>

  • Formula: \bold{Li_{2}O}
  • Ion Charges: \bold{Li~1+,Li~1+,O~2-}

<u>4.) Lithium and Nitrogen:</u>

  • Formula: \bold{Li_{3}N}
  • Ion Charges: \bold{Li~1+,Li~1+,Li~1+,N~3-}

<u>5.) Magnesium and Sulfur:</u>

  • Formula: \bold{MgS}
  • Ion Charges: \bold{Mg~2+,S~2-}

<u>6.) Magnesium and Chlorine:</u>

  • Formula: \bold{MgCl_2}
  • Ion Charges: \bold{Mg~2+,Cl~1-,Cl~1-}

<u>7.) Magnesium and Oxygen:</u>

  • Formula: \bold{MgO}
  • Ion Charges: \bold{Mg~2+,O~2-}

<u>8.) Magnesium and Nitrogen: </u>

  • Formula: \bold{Mg_3N_2}
  • Ion Charges: \bold{Mg~2+,Mg~2+,Mg~2+,N~3-,N~3-}
<h2 /><h2><u>Explanation:</u></h2>

______________________________

<h3><u>Lithium and Sulfur: </u></h3>
  • In order to make Lithium Sulfide, There must be 2 Lithium and 1 Sulfur. You transfer the electrons from both Lithium's to the Sulfur.
<h3 /><h3><u>Lithium and Chlorine:</u></h3>
  • In order to make Lithium Chloride, There must be 2 Lithium and 2 Chlorine. You transfer the electrons from both Lithium's to the Chlorines, (One electron for each chlorine.)
<h3 /><h3><u>Lithium and Oxygen:</u></h3>
  • In order to make Lithium Oxide, There must be 2 Lithium and 1 Oxygen. You transfer the electrons from both Lithium to Oxygen.  
<h3 /><h3><u>Lithium and Nitrogen:</u></h3>
  • In order to make Lithium Nitride, There must be 3 Lithium and 1 Nitrogen. You transfer the electrons from all 3 Lithium to Nitrogen.  
<h3 /><h3><u>Magnesium and Sulfur:</u></h3>
  • In order to make Magnesium Sulfide, There must be 1 Magnesium and 1 Sulfur. You transfer the both electrons from Magnesium to Sulfur.
<h3 /><h3><u>Magnesium and Chlorine:</u></h3>
  • In order to make Magnesium Chloride, There must be 1 Magnesium and 2 Chlorine. You transfer on electron to each Chlorine.
<h3 /><h3><u>Magnesium and Oxygen:</u></h3>
  • In order to make Magnesium Oxide, There must be 1 Magnesium and 1 Oxygen. You transfer both electrons from Magnesium to Oxygen.
<h3 /><h3><u>Magnesium and Nitrogen:</u></h3>
  • In order to make Magnesium Nitride, There must be 3 Magnesium and 2 Nitrogen. You transfer 3 electrons from Magnesium to Nitrogen.
<h3 />

______________________________

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1.Principle quantum number(n)

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
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Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

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