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Brut [27]
2 years ago
7

Oliver builds a circuit connecting a light bulb to a battery with wires, leaving a gap in one of the wires. He places several ob

jects across the gap to close the loop. He wants to see which objects allow electricity to flow and turn on the light bulb. Why do some materials allow electricity to flow through while others do not?
Plz help

A. Electricity will flow if the atoms in the material are bound tightly to each other.

B. Electricity will flow if the atoms in the material are bound loosely to each other.

C. Electricity will flow if the electrons are bound tightly to their atoms in the material.

D. Electricity will flow if the electrons are bound loosely to their atoms in the material.
Chemistry
1 answer:
kipiarov [429]2 years ago
7 0

Answer:

It is c

Explanation:

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a 1.25g sample of ore containing iron pyrite (FeS2) was pulverized and ignited in air, converting the FeS2 to Fe2O3 and SO2(g).
svp [43]

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:

  • The molar mass of Fe₂O₃ is 159.6 g/mol
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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:

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4 0
3 years ago
A sample was then prepared containing 14.00 mL of coffee and 8.00 mL of 4.80 ppm Li , and diluted to a 50.00 mL total volume. Th
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Answer:

The right answer is "[Na^+]=5.57 \ ppm".

Explanation:

The given values are:

\frac{A_s}{A_{coffee}} =\frac{0.840}{1.000}

According to the question,

The concentration of standard will be:

=  \frac{7.80}{50.00}\times 4.80

=  0.156\times 4.80

=  0.748 \ ppm

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⇒  \frac{A_{coffee}}{C_{coffee}} =F\times \frac{A_s}{C_s}

or,

⇒  C_{coffee}=\frac{A_{coffee}}{A_s}\times \frac{C_s}{F}

On substituting the values, we get

⇒               =\frac{1.000}{0.840}\times \frac{0.748}{1.68}

⇒               =1.191\times 1.3104

⇒               =1.561 \ ppm

hence,

In unknown sample, the concentration of coffee will be:

=  1.561\times \frac{50}{14}

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