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3 years ago

A percent composition analysis yields 46.7% nitrogen and 53.3% oxygen. What is the empirical formula for the compound

1 answer:

6 0

Empirical formula is the simplest ratio of whole numbers of components in a compound
calculating for 100 g of compound
N O
mass 46.7 g 53.3 g
number of moles 46.7 g / 14 g/mol 53.3 g / 16 g/mol
= 3.33 = 3.33
divide by the least number of moles
3.33/ 3.33 = 1.00 3.33/3.33 = 0
number of atoms
N - 1
O - 1
therefore empirical formula is NO

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2 KClO2 (s) 0 2 KCl (s) + 3 02 (g)

Nonamiya [84]

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<h3>Further explanation</h3>

Given

Reaction

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

5 moles of KCIO3

Required

volume of O2

Solution

From the equation, mol O2 :

= 3/2 x moles KClO3

= 3/2 x 5 moles

= 7.5 moles

Assumed at STP( 1 mol = 22.4 L) :

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3 years ago

How many moles and numbers of ions of each type are present in the following aqueous solution? 63.1 mL of 1.85 M magnesium chlor

Alexxandr [17]

Answer:

We have 0.117 moles Mg^2+ and 0.234 moles Cl-

The number of Mg^2+ ions = 7.04 *10^22 ions

The number of Cl- ions = 1.4*10^23 ions

Explanation:

Step 1: Data given

Volume of magnesium chloride = 63.1 mL

Concentration of solution = 1.85 M

Step 2: Calculate moles MgCl2

Moles MgCl2 = concentration * volume

Moles MgCl2 = 1.85 M * 0.0631 L

Moles MgCl2 = 0.117 moles

Step 3: Calculate moles of ions

MgCl2 → Mg^2+ + 2Cl-

For 1 mol MgCl2 we have 1 mol Mg^2+ and 2 moles Cl-

For 0.117 moles MgCl2 we have 0.117 moles Mg^2+ and 2*0.117 = 0.234 moles Cl-

The number of Mg^2+ ions = 0.117 * 6.022 *10^23 = 7.04 *10^22 ions

The number of Cl- ions = 1.4*10^23 ions

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sp2606 [1]

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14. Which of the following is produced when propane (C3H8) is combusted completely? (2 points)

Ymorist [56]

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<h3>Further explanation </h3>

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If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

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3 years ago