Compare and contrast the number of representative particles and the mass of UF6 with the number of representative particles and
2 answers:
Answer:
Explanation:
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In this case, since the given masses of UF6 and CCl3CF3 are 225 g and 200 g respectively (missing information) if we want to compare and contrast the number of representative particles both the molar mass for those compounds (352.02g/mol and 187.3756g/mol respectively) and the Avogrado's number must be used as follows:
In such a way, one sees that since it is about different compounds having different molar masses, but in order to know the number of representative particles, the same Avogadro's number is used, the results are quite different and is evident just by the simple fact that they are different compounds.
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Answer : Both solutions contain molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain molecules.
Avogadro's Number is =
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules = ;
∴ Number of molecules = which will be =
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
The question is incomplete, here is the complete question:
Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.
Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.
<u>Answer:</u> The pressure equilibrium constant for the reaction is 32908.46
<u>Explanation:</u>
We are given
Initial partial pressure of ammonia = 2.2 atm
Initial partial pressure of oxygen gas = 2.4 atm
Equilibrium partial pressure of nitrogen gas = 0.99 atm
The chemical equation for the reaction of ammonia and oxygen gas follows:
<u>Initial:</u> 2.2 2.4
<u>At eqllm:</u> 2.2-4x 2.4-3x 2x 6x
Evaluating the value of 'x':
So, equilibrium partial pressure of ammonia = (2.2 - 4x) = [2.2 - 4(0.495)] = 0.22 atm
Equilibrium partial pressure of oxygen gas = (2.4 - 3x) = [2.4 - 3(0.495)] = 0.915 atm
Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm
The expression of for above equation follows:
Putting values in above equation, we get:
Hence, the pressure equilibrium constant for the reaction is 32908.46