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Citrus2011 [14]
3 years ago
12

Which of the following is true about an ionic compound?

Chemistry
2 answers:
Yuri [45]3 years ago
4 0

Answer:

The true ones are

B. It is held together by ionic bonds.

C. It is composed of anions and cations

Explanation:

A is not true because even though it is true that a salt is an ionic compound, some ionic compounds are not salts. Some are metal oxides, some are acids, bases and a variety of different compounds which, though ionic are not salts

Genrish500 [490]3 years ago
3 0
 <span>D. all of the above because ionic compound is a salt. it is also held together by ionic bonds. it has anions and cations </span>
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Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3
AveGali [126]

Answer : The percent yield of CO_2 is, 68.4 %

Solution : Given,

Mass of CH_4 = 0.16 g

Mass of O_2 = 0.84 g

Molar mass of CH_4 = 16 g/mole

Molar mass of O_2 = 32 g/mole

Molar mass of CO_2 = 44 g/mole

First we have to calculate the moles of CH_4 and O_2.

\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 0.026 moles of O_2 react with \frac{0.026}{2}=0.013 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO_2

From the reaction, we conclude that

As, 2 mole of O_2 react to give 1 mole of CO_2

So, 0.026 moles of O_2 react to give \frac{0.026}{2}=0.013 moles of CO_2

Now we have to calculate the mass of CO_2

\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2

\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g

Theoretical yield of CO_2 = 0.572 g

Experimental yield of CO_2 = 0.391 g

Now we have to calculate the percent yield of CO_2

\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100

\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%

Therefore, the percent yield of CO_2 is, 68.4 %

6 0
3 years ago
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