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vichka [17]
3 years ago
5

Another metal phosphate is cobalt phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net io

nic equation including phases for CoPO4(s) dissolving in H3O^+(aq)?
Chemistry
2 answers:
ryzh [129]3 years ago
8 0
We are given with
Cobalt phosphate - CoPO4

We are asked for the net ionic equation for the phosphate dissolving in H3O+

The net ionic equation is
CoPO4 (s) + H3O+ (aq) ----->  HPO42- (aq) + Co3+ (aq) + H2O *(l)
Sidana [21]3 years ago
7 0

Answer: The equation is given below.

Explanation:

Calcium phosphate does not dissolve in water but in acid it does.Similarly cobalt phosphate is also soluble in acidic solution as it get dispersed in the acid solution.

CoPO_4(s)+H_3O^(aq)+\rightarrow HPO_{4}^{2-}(aq)+Co^{3+}(aq)+H_2O(l)

When cobalt phosphate dissolves in acidic solution as it get strongly attracted by the acid molecules and dissociates into cobalt(III) ions, hydrogen phosphate and water molecule.

Hence, the net ionic equation for cobalt phosphate in hydronium ion is given above.

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combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

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