Input interpretation
1 cm^3 of ammonia-15 N
Basic properties for 1 cm^3
mass | 0.814 grams 814 mg (milligrams) 8.14×10^-4 kg (kilograms) molar amount | 45.2 mmol (millimoles) 0.0452 mol (moles) volume | 1 mL (milliliter) 1000 µL (microliters) 0.001 L (liters) 1×10^-6 m^3 (cubic meters) 0.203 tsp (teaspoons) 0.00106 quarts equivalents | 0.0452 eq (equivalents) (at STP)
Corresponding quantities
sphere radius | 6.204 mm (millimeters) side of a cube | 10 mm (millimeters)
Mass composition for 1 cm^3
N (nitrogen) | 6.774 mg (0.8%) H (hydrogen) | 1.366 mg (0.2%)
Mass composition for 1 cm^3
Lewis structure
Draw the Lewis structure of ammonia-15 N. Start by drawing the overall structure of the molecule: Count the total valence electrons of the hydrogen (n_H, val = 1) and nitrogen (n_N, val = 5) atoms: 3 n_H, val + n_N, val = 8 Calculate the number of electrons needed to completely fill the valence shells for hydrogen (n_H, full = 2) and nitrogen (n_N, full = 8): 3 n_H, full + n_N, full = 14 Subtracting these two numbers shows that 14 - 8 = 6 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 3 bonds and hence 6 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 8 - 6 = 2 electrons left to draw: Answer: | |
Chemical names and formulas
formula | ^15NH_3 Hill formula | H_3(15N) name | ammonia-15 N IUPAC name | azane
Substance properties
molar mass | 18.024 g/mol phase | gas (at STP) melting point | -78 °C boiling point | -33 °C density | 0.814 g/cm^3 (at 25 °C)
Units
