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elena-s [515]
3 years ago
11

Certain compound contains 7.3% carbon, 4.5% hydrogen, 36.4% oxygen, and 31.8% nitrogen. It’s reality molecular mass is 176.0. Fi

nd its empirical and molecular formula
Chemistry
1 answer:
erastovalidia [21]3 years ago
6 0

Answer:

The answer to your question is:

Explanation:

Data

carbon        7.3%          =     7.3g

hydrogen    4.5%         =      4.5g

oxygen       36.4%         =     36.4 g

nitrogen     31.8%         =     31.8 g

Now

For carbon

                    12 g --------------------1 mol

                    7.3 g     -------------     x

                       x = 7.3/12 = 0.608 mol

For hydrogen

                 1 g   --------------------  1 mol

                 4.5 g  ------------------    x

                   x = 4.5 mol

For oxygen

             16 g ------------------- 1 mol

             36.4 g ----------------    x

             x = 2.28 mol

For nitrogen

              14 g   ----------------   1 mol

              31.8 g ---------------    x

             x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon              0.608/0.608 = 1

hydrogen           4.5/ 0.608 = 7.4

oxygen              2.28/0.608 = 3.75

nitrogen             2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

     

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8 0
2 years ago
The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.
Vikki [24]

Answer :  The solubility of this compound in g/L is 565.414\times 10^{-6}g/L.

Solution : Given,

K_{sp}=2.42\times 10^{-11}

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The balanced equilibrium reaction is,

                      MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3

At equilibrium                         s       s

The expression for solubility constant is,

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Now put the given values in this expression, we get

2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L

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3 years ago
Consider the SCl2 molecule. (a) What is the electron config- uration of an isolated S atom? (b) What is the electron con- figura
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Answer:

(a) 1s² 2s² 2p⁶ 3s² 3p⁴

(b) 1s² 2s² 2p⁶ 3s² 3p⁵

(c) sp³

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Explanation:

<em>Consider the SCl₂ molecule. </em>

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<em>(b) What is the electron configuration of an isolated Cl atom? </em>

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<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>

SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.

<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>

No valence orbital remains unhybridized.

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