Answer:
9.62moles of NH₃
Explanation:
Given parameters:
Number of moles of nitrogen = 4.81moles
Unknown:
Number of moles of NH₃ = ?
Solution:
To solve this problem, we need to establish a balanced reaction equation:
N₂ + 3H₂ → 2NH₃
Now, we can solve the problem by working from the known to the unknown.
1 mole of N₂ produced 2 moles of NH₃
4.81moles of N₂ will produce 2 x 4.81 = 9.62moles of NH₃
Answer:
Mass = 4.44 g
Explanation:
Given data:
Number of moles of Ne = 0.220 mol
Mass in gram = ?
Solution:
Formula:
Number of moles = mass/ molar mass
Molar mass of Ne = 20.2 g/mol
by putting values,
0.220 mol = mass/ 20.2 g/mol
Mass = 0.220 mol × 20.2 g/mol
Mass = 4.44 g
First, calculate the number of moles of sodium present with the given mass,
31.5 g of sodium x (1 mol sodium/ 23 g sodium) = 1.37 mol sodium
It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.
mols of H2 = (1.37 mols sodium)(1 mol H2/ 2 mols sodium)
mols of H2 = 0.685 mols H2
Then, at STP, 1 mol of gas = 22.4 L.
volume of H2 = (0.685 mols H2)(22.4 L / 1 mol)
volume of H2 = 15.34 L
Answer: 15.34 L
Answer:
Electrospray ionization
Explanation:
Electrospray ionization Is a soft ionization technique used in producing ions from macromolecules, it is employed especially in spectrometry where high voltage is used to create an aerosol from a liquid.It can be employed in Knowing molecular weights of molecules and biological macromolecules such as Peptides and proteins. Therefore, electrospray ionization is the method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer
