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Tcecarenko [31]
3 years ago
12

The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces

large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide ( NaN3 ) into solid sodium and gaseous dinitrogen. 2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm . Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.
Chemistry
1 answer:
Oxana [17]3 years ago
5 0

Answer:

1. NaN₃(s) → Na(s) + 1.5 N₂(g)

2. 79.3g

Explanation:

<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>

NaN₃(s) → Na(s) + 1.5 N₂(g)

<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>

First, we have to calculate the moles of N₂ from the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm.(43.0L)}{(0.08206atm.L/mol.K).286.2K} =1.83mol

The moles of NaN₃ are:

1.83molN_{2}.\frac{1molNaN_{3}}{1.5molN_{2}} =1.22molNaN_{3}

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

1.22mol.\frac{65.01g}{mol} =79.3g

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Liono4ka [1.6K]

Answer:volume

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Explanation:

7 0
3 years ago
The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for
tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. K(s)+O_2(g)\rightarrow KO_2(s)

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g)

Answer: A. K(s)+O_2(g)\rightarrow KO_2(s) : decreases

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g) : decreases

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g): no change

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g) : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction K(s)+O_2(g)\rightarrow KO_2(s), as gaseous reactant is changed to solid product, entropy decreases.

In reaction CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g), as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g), as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction  N_2O_2(g)\rightarrow 2NO(g)+O_2(g) , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0
3 years ago
Help please I need it bad
natta225 [31]
= 9.1 × 10^6
(scientific notation)

= 9.1e6
(scientific e notation)

= 9.1 × 10^6
(engineering notation)
(million; prefix mega- (M))

= 9100000
<span>(real number)</span>
5 0
3 years ago
Draw the structural formula of the major product of the reaction of (S)-2,2,3-trimethyloxirane with MeOH, H . Use the wedge/hash
Katarina [22]

Answer:

(S)-3-methoxy-3-methylbutan-2-ol

Explanation:

In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (CH_3OH) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.

See figure 1

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8 0
3 years ago
A 44.8 g rhodium sample contains how many rhodium atoms
Tresset [83]

Answer:

2.63*10^23

Explanation:

1 mol rhodium = 102.91

44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =

2.63*10^23

6 0
3 years ago
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