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ladessa [460]
3 years ago
5

An unknown compound is processed using elemental analysis and found to contain 117.4 g of platinum, 28.91 g of carbon, and 33.71

g of nitrogen. How many moles of platinum are present in the compound? Need help ASAP
Chemistry
1 answer:
VLD [36.1K]3 years ago
4 0

Answer:

1 mole of platinum (Pt).

Explanation:

From the question given above, the following data were obtained:

Platinum (Pt) = 117.4 g

Carbon (C) = 28.91 g

Nitrogen (N) = 33.71 g

Number of mole of platinum =?

To obtain the number of mole of platinum present in the compound, we shall determine the empirical formula for the compound. The empirical formula for the compound can be obtained as follow:

Pt = 117.4 g

C = 28.91 g

N = 33.71 g

Divide by their molar mass

Pt = 117.4 /195 = 0.602

C = 28.91 /12 = 2.409

N = 33.71 /14 = 2.408

Divide by the smallest

Pt = 0.602 /0.602 = 1

C = 2.409 /0.602 = 4

N = 2.408 / 0.602 = 4

Therefore, the empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the empirical formula for the compound (i.e Pt(CN)₄) we can see that the compound contains 1 mole of platinum (Pt).

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Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" 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 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The <u>reaction that takes place is</u>:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. We use PV=nRT to <u>calculate the moles of N₂ that were produced</u>.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

  • 1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 K
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Now we <u>convert N₂ moles to NaN₃ moles</u>:

  • 0.334 mol N₂ * \frac{2molNaN_{3}}{3molN_2} = 0.223 mol NaN₃

Finally we <u>convert NaN₃ moles to grams</u>, using its molar mass:

  • 0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃

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Answer:

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Explanation:

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Answer:

Explanation:

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Answer:

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Explanation:

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