How many sodium atoms are there in 6.0 g of #Na_3N#?

Answer 1

#"Moles of sodium"# #=# #3xx"moles of sodium nitride"#

#"Moles of sodium nitride"# #=# #(6.0*g)/(82.99*g*mol^-1)#

Consequently, the quantity of sodium atoms (or sodium ions, if less):

#(6.0*g)/(82.99*g*mol^-1)xxN_Axx3#, where #N_A=6.022xx10^23*mol^-1#
i.e. #3xx(6.0*g)/(82.99*g*mol^-1)xx6.022xx10^23*mol^-1#. We get an actual number, without units, as required.
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Answer 2

In order to determine how many sodium atoms are in 6.0 g of Na3N, we must first determine the molar mass of Na3N, which is 65 grams per mole (22.99 g/mol for Na multiplied by 3, plus 14.01 g/mol for N). Next, we convert the mass of Na3N to moles using Avogadro's number, which is roughly (6.022 \times 10^{23}) atoms/mol. Finally, we multiply the number of moles by Avogadro's number to determine the number of sodium atoms: [6.0 \text{ g} \times \frac{1 \text{ mol}}{65 \text{ g}} \times 6.022 \times 10^{23} \text{ atoms/mol}] [= \frac{6.0 \times 6.022 \times 10^{23}}{65}] [= 5.54 \times 10^{23}] sodium atoms.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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