Urea is highly soluble in water. When it is allowed to dissolve in water in the presence of heat, it will yield ammonia and carbon dioxide. The reaction is shown below:
<span>NH2-CO-NH2 + H2O </span>⇒ 2 NH3 + CO2
As you can observe in the stoichiometric equations, 1 molecule of water can dissolve with 1 mole of urea.
Empirical formula is the simplest ratio of components making up a compound.
The percentage composition of each element has been given
therefore the mass present of each element in 100 g of compound is
B N H
mass 40.28 g 52.20 g 7.53 g
number of moles
40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol
= 3.662 mol = 3.729 mol = 7.53 mol
divide the number of moles by the least number of moles, that is 3.662
3.662 / 3.662 3.729 / 3.662 7.53 / 3.662
= 1.000 = 1.018 = 2.056
the ratio of the elements after rounding off to the nearest whole number is
B : N : H = 1 : 1 : 2
therefore empirical formula for the compound is B₁N₁H₂
that can be written as BNH₂
The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g
<h3>Conversion scale </h3>
1 u = 1.66×10¯²⁴ g
<h3>How to determine the mass of hydrogen atoms </h3>
- Mass of Hydrogen (u) = 54 u
- Mass of Hydrogen (g) =?
1 u = 1.66×10¯²⁴ g
Therefore
54 u = 54 × 1.66×10¯²⁴ g
54 u = 89.64×10¯²⁴ g
Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g
Learn more about conversion:
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Answer:
the mass number of the atom or ion is 6
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