<span>Molar mass is the mass of
one mole of a substance, it can be a chemical element or a compound. It is a
characteristic of each pure substance. We calculate it by adding up all of the masses of the atoms involved in the compound. We calculate as follows:
atomic mass total mass
C 17 12.01 g/mol 204.17 g/mol
H 19 1.01 g/mol 19.19 g/mol
N 1 14.00 g/mol 14.00 g/mol
O 3 16.00 g/mol 48.00 g/mol
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Molar mass = 285.36 g/mol
</span><span>What is the mass of 6.02 x 10^24 molecules of morphine?
</span>6.02 x 10^24 molecules ( 1 mol / 6.02x10^23 molecules) ( 285.36 g/mol) = 2853.6 g morphine
For what exactly ? just for fun or for educational purposes?
Answer:
156 g
Explanation:
Let's consider the following reaction.
2 NaN₃(s) → 2 Na(s) + 3 N₂
(g)
We can find the moles of N₂ using the ideal gas equation.
P × V = n × R × T
1.50 atm × 60.0 L = n × (0.08206 atm.L/mol.K) × 305 K
n = 3.60 mol
The molar ratio of N₂ to NaN₃ is 3:2. The moles of NaN₃ are:
3.60 mol N₂ × (2 mol NaN₃ / 3 mol N₂) = 2.40 mol NaN₃
The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:
2.40 mol × 65.01 g/mol = 156 g
Answer:
Barium has a larger atomic radius because it has more energy levels and a larger shielding effect. Silicon has a smaller atomic radius because it has a stronger nuclear charge. ... Aluminum has a higher ionization energy than indium because there are fewer energy levels so it is more challenging to remove an electron.
Explanation:
Answer:
% recovery
MP range of product
mass of product
Explanation:
Liquid–liquid extraction (LLE) is a process of transferring one (or more) solute(s) which are present in a feed solution to another immiscible liquid (solvent). The other solvent that becomes enriched in the target solute(s) is called extract. The original feed solution that is depleted in solute(s) is subsequently referred to as the raffinate.
This method is used to purify compounds and separate mixtures of compounds. This is very important when we want to isolate a product from a reaction mixture.
The percent recovery is the amount of solute that is transferred to the extract. This is the most important data to be recorded in an LLE experiment.
The melting point range necessarily helps us to identify the product and the mass of solid tells us the quantity of the solid obtained after extraction.
