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I am Lyosha [343]
3 years ago
8

The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so th

at the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is: 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq).
When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 ∘C was 0.961 L at a total pressure of 748 mmHg .
Chemistry
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

A penny dissolves in hydrochloric acid if the copper coating is filed down in several spot... ... When The Zinc In A Certain Penny Dissolves, The Total Volume Of Gas ... is filed down in several spots (so that the hydrochloric acid can reach the zinc). The reaction between the acid and the zinc 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq) .

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You have 16.0 g of some compound and you perform an experiment to remove all of the oxygen, 11.2 g of iron is left. What is the
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The empirical formula of this compound is equal to Fe_{2}O_3.

<h3>Empirical formula</h3>

To calculate the empirical formula of a compound, it is necessary to know the number of moles present.

Therefore, we will use the molar mass of iron and oxygen to find the amount of moles, so that:

MM_O = 16g/mol\\MM_Fe = 55.8g/mol

                                        MM = \frac{m}{mol}

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                                            16 = \frac{4.8}{mol}

                                            mol = 0.3

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                                               55.8 = \frac{11.2}{mol}

                                               mol = 0.2

Finally, as the empirical formula is composed of integers numbers of moles, just multiply the values ​​by the smallest common factor to transform into an integer, so that:

                                       O =>  0.3 \times 10 = 3moles

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Learn more about empirical formula in: brainly.com/question/1363167

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3 years ago
A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro
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Answer:

The specific heat of the metal is 0.34 J/g°C

Explanation:

Step 1: Data given

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Mass of the water = 25.0 grams

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Initial temperature of the water = 22.5 °C

Final temperature = 25 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Qlost = Qgained

Q = m*c*ΔT

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m(metal) *c(metal)* ΔT(metal) = -m(water) * c(water) *ΔT(water)

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⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C

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