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Firlakuza [10]
3 years ago
12

6. If 0.500 mol NaN3 react, what mass in grams of nitrogen will result?

Chemistry
1 answer:
lions [1.4K]3 years ago
7 0

Answer:

21 g of N₂ are produced by the decomposition

Explanation:

The reaction is: 2 NaN3 → 2 Na + 3 N2

2 moles of sodium nitride decompose in order to produce 2 moles of Na and 3 moles of nitrogen gas.

According to stoichiometry, ratio is 2:3. Therefore we say,

2 moles of nitride can produce 3 moles of N₂

Then, 0.5 moles of NaN₃ will produce (0.5 . 3) / 2 = 0.75 moles of N₂

We convert the moles to mass, to find the answer

0.75 mol . 28 g / 1 mol = 21 g

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Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

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Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

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The expression for solubility constant for this reaction will be,

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Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

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Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

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3 years ago
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3 years ago
Read 2 more answers
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