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netineya [11]
2 years ago
7

When a pot of water sits on a hot stove, the metal pot is heated, and then transfers its heat to the water. A student claims tha

t conduction is the only type of heat transfer occurring. Which statement explains why the student's idea is incorrect?
a. Conduction cannot occur between two different phases of matter.
b. The water transfers heat within the pot by convection.
c. Water cannot be heated by electromagnetic radiation.
d. Metal is a poor conductor of heat.
Chemistry
2 answers:
finlep [7]2 years ago
3 0

When a pot of water sits on a hot stove, the metal pot is heated, and then transfers its heat to the water. A student claims that conduction is the only type of heat transfer occurring. The statement that explains why the student’s idea is incorrect Conduction cannot occur between two different phases of matter. The answer is letter A.

MissTica2 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

When we heat a fluid then the cooler material which is more denser in nature will sink at the bottom whereas the hot material which is less denser in nature will rise at the top. This flow of material induces currents and the process is known as convection.

For example, when a metal pot of water is heated then heat will be transferred from pot to the water represents convection process.

Whereas when two objects are touching each other and both of them have different temperature then heat will flow from warmer object to cooler object. This process is known as conduction.

Thus, we can conclude that the statement conduction cannot occur between two different phases of matter explains why the student's idea is incorrect.

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Answer:

28.9%

Explanation:

Let's consider the following balanced equation.

2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂

We can establish the following relations:

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The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

0.516gFe_{2}O_{3}.\frac{1molFe_{2}O_{3}}{159.6gFe_{2}O_{3}} .\frac{2molFeS_{2}}{1molFe_{2}O_{3}} .\frac{1molFe}{1molFeS_{2}} .\frac{55.84gFe}{1molFe} =0.361gFe

The percent of Fe in 1.25 g of the ore is:

\frac{0.361g}{1.25g} .100\%=28.9\%

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