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NARA [144]
3 years ago
13

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

the oxygen and gives a red binary compound that is 9.334% oxygen by mass. Strong heating drives off more oxygen and gives still another binary compound, which is only 7.168% oxygen by mass.
(a) Compute the mass of oxygen that is combined with 1.000 g of the metal in each of these three oxides.
(b) Assume that the empirical formula of the first compound is MO2 (where M represents the metal). Give the empirical formulas of the second and third compounds.
(c) Name the metal.
Chemistry
1 answer:
Leto [7]3 years ago
6 0

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

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Answer:

Explanation:

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b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%

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2 years ago
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Answer:

P_T=112.4torr

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

P_{H_2}=x_{H_2}*P_T

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P_T=\frac{P_{H_2}}{x_{H_2}}

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andrey2020 [161]

Answer:

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Explanation:

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Zanzabum

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Explanation:

Given parameter:

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Conversion           uL → mL

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liter is a unit of volume of a substance.

       micro - is 10⁻⁶

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The problem is converting from micro to milli:

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learn more:

Volume brainly.com/question/5055270

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2. acid hydrolysis.
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