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NaN3 also known as Sodium azide is used as an antibacterial agent and to inflate airbags.

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15y ago

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What is the empirical formula for sodium azide?

The empirical formula for sodium azide is NaN3.


What is the chemical formula for sodium azide?

NaN3


What is the total number of moles present in a 52.0 gram sample of NaN3 gram formula mass equals 65.0 mole?

To calculate the total number of moles in the 52.0 gram sample of NaN3, divide the given mass by the molar mass of NaN3. First, determine the molar mass of NaN3 by summing the atomic masses of its elements (sodium, nitrogen, and three times the atomic mass of nitrogen). Then, divide the mass of the sample by the molar mass of NaN3 to find the number of moles.


What is the main gas in an air bag that inflated using the NaN3 reaction?

NaN3 decomposed to form nitrogen gas (N2) and sodium. 2NaN3 --> 2Na + 3N2


What gas fills air bags?

Callod sodium acid ( NaN3 )


Is NaN3 covalent or ionic?

NaN3 (sodium azide) is an ionic compound. It is composed of sodium cations (Na+) and azide anions (N3-) which are held together by ionic bonds due to the transfer of electrons from sodium to azide.


What is the name of this compound NaN3?

The compound NaN3 is called sodium azide. It is commonly used as a propellant in airbags and as a reagent in organic synthesis. Sodium azide is highly toxic and should be handled with caution.


How many grams of NaN3 are required to from 6.00 g of N2 gas?

9.28 g


How is sodium azide decomposed?

The chemical equation is:2 NaN3 = 2 Na + 3N2


How many moles of N2 are produced by the decomposition of 2.88 mol of sodium azide?

Using the balanced chemical equation for the decomposition of sodium azide, 2NaN3 -> 2Na + 3N2, we can see that 2 moles of Na3N produce 3 moles of N2. Therefore, for 2.88 mol of NaN3, we would produce 2.88 mol * (3 mol N2 / 2 mol NaN3) = 4.32 moles of N2.


What is the chemical equation when decomposing sodium azide to sodium nitride?

NaN3 (sodium azide) decomposes into Na (sodium) and N2 (nitrogen gas). The chemical equation is: 2 NaN3(s) → 2 Na(s) + 3 N2(g)


How many grams of NaN3 are required to produce 13.0 ft3 of nitrogen gas if the gas has a density of 1.25 gL?

First, convert 13.0 ft3 to liters (1 ft3 is approximately 28.32 L). This gives approximately 368.96 L. Since the density of nitrogen gas is 1.25 g/L, the mass needed is 368.96 L * 1.25 g/L = 461.2 g. Finally, use the molar mass of NaN3 (65 g/mol) to convert grams to moles (461.2 g / 65 g/mol = 7.09 moles). Therefore, approximately 461.2 grams of NaN3 are required to produce 368.96 L of nitrogen gas.