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Assoli18 [71]
3 years ago
9

A 72.0-gram piece of metal at 96.0 °C is placed in 130.0 g of water in a calorimeter at 25.5 °C. The final temperature in the ca

lorimeter is 31.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.
Chemistry
1 answer:
tekilochka [14]3 years ago
6 0

Answer: The answer is S = 0.1528 cal/g °C

Explanation:

By the law of conservation of energy, energy is neither created nor destroyed.

So, energy lost by metal pieces is equal to the energy gained by water in the calorimeter.

Specific heat of water is 1 cal/g °C

⇒ heat energy  Q = mSΔT, where m = mass of a substance

                                                        S = specific heat

                                                        ΔT = change in temperature

Now, the heat lost by metal piece, Q = 72×S×(96-31)

                                                      = 4680×S cal

Heat gained by water, Q = 130×1×(31-25.5)

                                        = 715 cal

⇒ 4680×S = 715.

⇒ S = 0.1528 cal/g °C.

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molecules of i2 are produced by the reaction of 0.4321 mol of cucl2 according to the following equation. 2 cucl2 4 ki → 2 cui 4
Juli2301 [7.4K]

\: 1.237 \times 10^{23}   molecules \: of \:I _{2} \: are

are\: produced \: and \: 53.74 \: g \: of \: I _{2} \: are

produced from the reaction.

The overall balanced equation for the reaction is,

2CuCl_{2} +   KI→2CuI + 4KCl  + I_{2}

Number  \: of  \: moles  \: ofCuCl_{2}=0.4235 \: mol

2 \: moles \: of \: CuCl_{2} \: produce \: \: 1 \: mole \:

of \:  I_{2}.

2 \: mole \: of \: CuCl_{2}  = 1 \: mole \: of \:  I_{2}

1 \: mole \: of \: I_{2} \:  = 6.022 \times 10 ^{23}  \: molecules \: of  I_{2}

So, the number of molecules produced in the reaction of

I_{2} \: are

=  \frac{0.4235 \times 6.022 \times 10 ^{23} }{2 }

= 1.237 \times 10^{23}  \: molecules \:

1.237 \times 10^{23}  \: molecules \: of \: I_{2} \: are \:

produced from the reaction.

Mass  \: of \:   I _{2} \: produced \: are \: is ,

=  \frac{0.4235 \times 253.8}{2}

= 53.74 \: g

53.74 \: g \: of \:  I _{2} \: produced \: in \: the \:

reaction.

Therefore , \: 1.237 \times 10^{23}   molecules

of  \: I_{2} \: are\: produced \: and \: 53.74 \: g \: of

are produced in the reaction.

To know more about moles, refer to the below link:

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7 0
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Discussion topic
FromTheMoon [43]

Answer: Neither scientist discovered “ Avogadro's number” in the form we use it today (6.02 x 10 23). Still, there's a controversy ...

Explanation: hope this helps

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How do you draw a electron dot diagram
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Imagine a square. In an electron dot diagram, the elements symbol is in the center. It represents the nucleus and the inner electrons. The valence electrons are drawn as dots on the sides of the square. These electrons are distributed one on each side, and then paired up. For example, chlorine is in period 17 and has 7 valence electrons. "Cl" would go in the center of the square, and then there would would be two dots on three of the square's sides, and one dot on the other. However, the sides of the square are imaginary, don't draw them.
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