Gold (III) nitrate in an aqueous solution is hydrolyzed with formation of gold (III) metahydroxide.
Au(NO₃)₃ → Au³⁺(aq) + 3NO₃⁻(aq)
Au³⁺ + H₂O ⇄ AuOH²⁺ + H⁺
AuOH²⁺ + H₂O ⇄ Au(OH)₂⁺ + H⁺
Au(OH)₂⁺ + H₂O → AuOOH·H₂O(s) + H⁺
Au(NO₃)₃(aq) + 2H₂O(l) = AuOOH(s) + 3HNO₃(aq)
An oxide of nitrogen contains 30.45 mass % N, if the molar mass is 90± 5 g/mol the molecular formula is N₂O₄.
<h3>What is molar mass?</h3>
The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. A substance's molar mass is one of its properties. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.
<h3>How to find the molecular formula?</h3>
The whole-number multiple is defined as follows.
Whole-number multiple = ![\frac{molar mass (g/mol)}{empirical formula mass (g/mol)}](https://tex.z-dn.net/?f=%5Cfrac%7Bmolar%20mass%20%28g%2Fmol%29%7D%7Bempirical%20formula%20mass%20%28g%2Fmol%29%7D)
The empirical formula mass is shown below.
Mw of empirical formula = Mw of N+ 2 x (Mw of O)
= 14.01 g/mol + 2 x (16.00 g/mol)
= 46.01 g/mol
With the given molar mass or the molecular formula mass, we can get the whole-number multiple for the compound.
Whole-number multiple =
≈ 2
Multiplying the subscripts of NO2 by 2, the molecular formula is N(1x2)O(2x2)= N2O4.
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Answer:
Explanation:
To determine the molecular formula of the compound, the empirical formula must be determined first. To determine the empirical formula, the percentage of each constituent is divided by its molar mass. This is shown below
Carbon = 60/12 = 5
Oxygen = 32/16 = 2
Hydrogen = 8/1 = 8
The next step is to divide each ratio by the smallest value. The smallest value is 2. It becomes
Carbon = 5/2 = 2.5
It is approximated to 3
Oxygen = 2/2 = 1
Hydrogen = 8/2 = 4
Therefore, the empirical formula is
C3H4O
From the given relative molecular mass of the compound, the molecular formula can be determined
Answer:
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Explanation:
Answer:
6.142 moles of NaCl
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2AlCl3 + 3Na2S —> Al2S3 + 6NaCl
Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:
Mass mass of Na2S = 78.048g/mol
Mass of Na2S = 239.7g
Number of mole Na2S =..?
Mole = Mass /Molar Mass
Number of mole Na2S = 239.7/78.048 = 3.071 moles
Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:
From the balanced equation above,
3 moles of Na2S reacted to produce 6 moles of NaCl.
Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl