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neonofarm [45]
3 years ago
8

Vitamin C contains only carbon, hydrogen, and oxygen. When a 1.00 g was combusted, 1.4991 g of CO2 and 0.4092 g of H2O were obta

ined. Determine the empirical formula for this vitamin.
Chemistry
1 answer:
Alina [70]3 years ago
5 0

The empirical formula for this vitamin : C₃H₄O₃

<h3>Further explanation   </h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

  • Determine the mass ratio of the constituent elements of the compound.  
  • Determine the mole ratio by dividing the percentage by the atomic mass

Mass of C in CO₂ :(MW C = 12 g/mol, CO₂=44 g/mol)

\tt \dfrac{12}{44}\times 1.4991=0.409~g

Mass of H in H₂O :(MW H = 1 g/mol, H₂O = 18 g/mol)

\tt \dfrac{2.1}{18}\times 0.4092=0.0455~g

Mass O = Mass sample - (mass C + mass H) :

\tt 1-(0.409+0.0455)=0.5455~g

mol ratio C : H : O =

\tt \dfrac{0.409}{12}\div \dfrac{0.0455}{1}\div \dfrac{0.5455}{16}\\\\0.0341\div 0.0455\div 0.0341\rightarrow 1\div 1.33\div 1=3\div 4\div 3

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Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

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4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

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7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

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Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
3 years ago
3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O
liubo4ka [24]

24.25 moles of NO can be produced using 97 moles of HNO3.

<h3>What is balanced chemical equation?</h3>

Equal numbers of atoms from various elements are present in both the reactants and the products in balanced chemical equations. Varied elements' atom counts in the reactants and products of unbalanced chemical equations are different.

3 Cu + 8HNO3 g → 3 Cu(NO3)2 + 2 NO + 4 H2O

The number of moles consumed can be calculated using comparing with coefficients in the balanced reaction .

So , from above eq we get that 8 moles of HNO3 are consumed to make 2 moles of NO.

⇒  8 HNO3⇔2 NO

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This means that for each mole of HNO3 produces 1/4 moles of NO.

So , for 97 moles of HNO3 , \frac{1}{4}  *97 moles of NO can be made,

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Lean more about balanced reactions here brainly.com/question/26694427

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