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AveGali [126]
3 years ago
5

Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. Th

is causes sodium azide (NaN3) to decompose explosively according to the following reaction. 2 NaN3(s) --> 2 Na(s) 3 N2(g) What mass in grams of NaN3(s) must be reacted in order to inflate an air bag to 79.5 L at STP
Chemistry
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

154 g

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

79.5 L × 1 mol/22.4 L = 3.55 mol

Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.

Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.37 mol × 65.01 g/mol = 154 g

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A gas phase mixture of H2 and N2 has a total pressure of 784 torr with an H2 partial pressure of 124 torr. What mass of N2 gas i
olchik [2.2K]

The mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g

We'll begin by calculating the partial pressure of N₂. This can be obtained as follow:

Total pressure = 784 torr

Partial pressure of H₂ = 124 torr

<h3>Partial pressure of N₂ =?</h3>

Total pressure = Partial pressure of H₂ + Partial pressure of N₂

784 = 124 + Partial pressure of N₂

Collect like terms

Partial pressure of N₂ = 784 – 124

<h3>Partial pressure of N₂ = 660 Torr</h3>

  • Next, we shall determine the number of mole of N₂ in the mixture.

Pressure of N₂ (P) = 660 Torr

Volume (V) = 2 L

Temperature (T) = 298 K

Gas constant (R) = 62.364 L.Torr/Kmol

<h3>Number of mole of N₂ (n) =? </h3>

PV = nRT

660 × 2 = n × 62.364 × 298

1320 = n × 18584.472

Divide both side by 18584.472

n = 1320 / 18584.472

<h3>n = 0.071 mole</h3>

  • Finally, we shall determine the mass of N₂ in the mixture.

Mole of N₂ = 0.071 mole

Molar mass of N₂ = 2 × 14 = 28 g/mol

<h3>Mass of N₂ =? </h3>

Mass = mole × molar mass

Mass of N₂ = 0.071 × 28

<h3>Mass of N₂ = 1.988 g</h3>

Therefore, the mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g

Learn more: brainly.com/question/20853110

4 0
2 years ago
Draw the Lewis structure for the polyatomic hydronium H3O cation. Be sure to include all resonance structures that
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Answer:

 Lewis structure of Hydronium ion is shown below :                          

Explanation:

Lewis structure : It is a representation of valence electrons on the atoms in a molecule

Here , Hydronium ion is given , which contains 1 atom of oxygen and 3 atoms of hydrogen .

Oxygen has a total of 6 valence electrons and hydrogen contains 1 valence electron .

Oxygen share its 3 valence electrons with 3 hydrogen atoms and left with 3 valence electrons. From these three valence  electrons of oxygen atom  two electrons will be shown as a pair of electrons on oxygen atom but a single electron can not be shown . So , to simplify this, one positive charge is shown overall .  

Resonance structure will be same as the hybrid structure because all  three atoms are same , that is hydrogen .

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What is the concentration of a solution formed by diluting 25.0 ml of a 3.2 M NaCl solution to 135.0 ml?
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Answer:

B) 0.59 M NaCl.

Explanation:

  • It is known that the no. of millimoles of NaCl before dilution = the no. of millimoles of NaCl after dilution.

∵ (MV) before dilution = (MV) after dilution.

<em>∴ M after dilution = (MV) before dilution / V after dilution </em>= (3.2 M)(25.0 mL)/(135.0 mL) = <em>0.5926 M ≅ 0.59 M.</em>

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