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ehidna [41]
3 years ago
5

What is the empirical formula for the following molecular formula: C10H5O2

Chemistry
1 answer:
Tju [1.3M]3 years ago
4 0

The empirical formula is the same as the molecular formula : C₁₀H₅O₂

<h3>Further explanation</h3>

Given

Molecular formula : C₁₀H₅O₂

Required

The empirical formula

Solution

The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.  

The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

<em>(EF)n=MF </em>

(EF)n = C₁₀H₅O₂

If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula

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write a balanced net ionic equation for the following reaction: BaCl2(aq) + H2SO4 (aq) -- BaSO4(s) + HCl (aq)
Fudgin [204]

The  balanced net  equation  for

BaCl2 (aq)  + H2SO4(aq) → BaSO4(s)  + HCl  (aq)  is

 Ba^2+(aq)  +SO4^2- → BaSO4 (s)

 <u><em>Explanation</em></u>

Ionic equation  is a chemical  equation in which  electrolytes  in aqueous  solution are written as dissociated ions.

<u>ionic equation is written using the below steps</u>

Step 1:  <em>write a balanced   molecular equation</em>

 BaCl2 (aq) +H2SO4 (aq)→ BaSO4(s)  +2HCl (aq)

Step 2:   <em>Break all soluble  electrolytes  in to ions</em>

=  Ba^2+ (aq) + 2Cl^-(aq)  + 2H^+(aq) + SO4^2-(aq)→ BaSO4(s)   + 2H^+(aq)  +2Cl^- (aq)


step 3:  <em>cancel the spectator  ions  in both side of equation ( ions which  do not take place in the reaction)</em>

<em> </em><em>    =</em> 2Cl^-  and  2H^+  ions

Step 4: <em>write the final net equation</em>

<em> Ba^2+(aq)  + SO4^2-(aq)→  BaSO4(s</em><em>)</em>

3 0
3 years ago
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What is the [h3o ] for a solution of seawater at 25°c that has a ph of 8.3?
Kipish [7]
PH of solution at 25ºC = 8.3

[ H_3O^+] = 10 ^{-pH}

{H_3O^+] = 10 ^{-8.3}

[H_3O^+] = 5.011*10^{-9}  M

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6 0
3 years ago
What is the percent composition of NaHCO3?
zhuklara [117]

Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

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