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2 years ago

Câu 12. Hiện tượng nào sau đây không phải là sự nóng chảy?

Chemistry

1 answer:

Alik [6]2 years ago

5 0

Mình nghĩ C là đáp án đúng

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Make an observation about the average height of the beanbag for each mass dropped. How does it compare with your calculated kine

Gekata [30.6K]

Answer:

the other person is correct

Explanation:

3 0

2 years ago

The two weather maps show a front moving across Texas in which of the following cities would a decrease in temperature be predic

Nadusha1986 [10]

Answer:

San Antonio and Dallas Texas

Explanation:

8 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

2 years ago

The average weather conditions of a region

Verdich [7]

Answer:

A) Climate

Hope this helps !

6 0

3 years ago

Anybody can help? - 20 points

strojnjashka [21]

Answer:

I think its carbon because of all the chemicals being used in that air.

7 0

3 years ago