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Irina-Kira [14]
3 years ago
14

The combustion of ethane ( C 2 H 6 ) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6 H 2 O (

g ) How many moles of CO 2 are produced when 5.90 mol of ethane is burned in an excess of oxygen?
Chemistry
1 answer:
raketka [301]3 years ago
4 0

Answer:

5.90 moles of ethane, C₂H₆, is burned in an excess of oxygen to produce 11.8 moles of CO₂.

Explanation:

To answer the question, we examine the chemical reaction as follows

2C₂H₆ ( g ) + 7O₂ ( g ) ⟶ 4CO₂ ( g ) + 6H₂O ( g )

From the chemical reaction it is seen that 2 moles of ethane, C₂H₆, is required to produce 4 moles of CO₂

This means that 1 mole of C₂H₆, is required to produce 2 moles of CO₂

Since we are asked to find out how many moles of CO 2 are produced when 5.90 moles of ethane is burned in an excess of oxygen and we already know that 1 mole of C₂H₆, is required to produce 2 moles of CO₂, we multiply both the 1 mole of C₂H₆ and the 2 moles of CO₂  by 5.90 to obtain;

(5.90 × 1 mole) of C₂H₆  is required to produce (5.90× 2 moles) of CO₂ or

5.90 moles of C₂H₆  is required to produce 11.8 moles of CO₂.

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For a theoretical yield of 2.97 g and percent yield of 93.4452%, calculate the actual yield for a chemical reaction. Answer in u
MariettaO [177]

Answer:

2.77532244 grams

Explanation:

Percent yeild=actual/theoretical * 100

Plug in the values

93.4452=actual/2.97 *100

0.934452*2.97=2.77532244

3 0
2 years ago
Binary compound of oxygen and an unknown element,X, has the formula XO2 and is 69.55 mass % oxygen, what is the weight of elemen
marissa [1.9K]

Answer:

Weight of element X = 14.01 gram

Explanation:

Mass percentage is calculated by using :

Percent =\frac{Mass\ of\ element}{Total\ mass}\times 100

O\ Percent=\frac{Mass\ of\ O}{Total\ mass\ of\ XO2}\times 100

Mass of O = 16.0 gram

let mass of X = X grams = ?

Mass of XO2 = mass of X + 2(mass of O)

Mass of XO2 = X + 2(16)

Mass of XO2 = X + 32

Total mass of O in the compound = 2(16) = 32 grams

Percent O = 69.55 %

Percent\ O =\frac{Mass\ of\ O}{Total\ mass\ of\ XO2}\times 100

69.55=\frac{32}{X+32}\times 100

\frac{69.55}{100}=\frac{32}{X+32}

0.6955=\frac{32}{X+32}

0.6955\times (X+32)=32

0.6955X+ 22.256=32

0.6955X=9.744

X=\frac{9.744}{0.6955}

X =14.01

weight of element X = 14.01 gram

14.01 is mass of nitrogen N

The correct formula of this element is NO2

6 0
3 years ago
When determining whether a chemical reaction has taken place you observe and look for several indicators. Which would be conside
Phoenix [80]

Answer:

B) Heat is given off.

Explanation:

When heat is given off, a chemical change must have occurred and such a reaction is termed an exothermic reaction.

A chemical change is one in which;

  • the process is not easily reversible
  • leads to the production of new kinds of matter
  • involves change in mass
  • requires a considerable amount of energy.

Most chemical reactions are accompanied by heat changes.

4 0
3 years ago
How many ml of a 4.0 M Mg(OH)2 are needed to make 750 ml of 0.2 M Mg(OH)2???
cupoosta [38]

Answer:

37.5 mL of a 4.0 M Mg(OH)₂ are needed to make 750 mL of 0.2 M Mg(OH)₂

Explanation:

Dilution is the reduction in concentration of a chemical in a solution. Dilution consists of lowering the amount of solute per unit volume of solution.

In other words, dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one, and consists of adding solvent to an existing solution. The quantity or mass of the solute is not changed but only that of the solvent.

In dilutions the expression is used:

Ci*Vi = Cf*Vf

where:

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

In this case:

  • Ci: 4 M
  • Vi: ?
  • Cf: 0.2 M
  • Vf: 750 mL

Replacing:

4 M* Vi= 0.2 M* 750 mL

Solving:

Vi=\frac{0.2 M*750 mL}{4 M}

Vi= 37.5 mL

<u><em>37.5 mL of a 4.0 M Mg(OH)₂ are needed to make 750 mL of 0.2 M Mg(OH)₂</em></u>

8 0
2 years ago
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Delicious77 [7]

Answer:

Heat, solutes and high temperature.

Explanation:

A supersaturated solution can be formed by dissolving solute more solute in solvent by increasing temperature of the solution. A supersaturated solution contains more quantity of solutes than can be dissolved in the solvent at room temperature. A solution may remain supersaturated until the solution has high temperature and when the temperature started lower, the extra dissolve solutes begin undissolved and remain suspended in the solution.

4 0
3 years ago
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