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3 years ago

If the water is initially at temperature 10.2 ∘C, how long will it take for its temperature to rise to 57.9 ∘C?

Chemistry

1 answer:

denpristay [2]3 years ago

7 0

Answer:

The complete part of the question is ; In the circuit in the figure, (Figure 1)a 20-ohm resistor sits inside 110 g of pure water that is surrounded by insulating Styrofoam. If the water is initially at temperature 10.2 ∘C, how long will it take for its temperature to rise to 57.9 ∘C?

Use 4190J/kg⋅C∘ as the heat capacity of water, and express your answer in seconds using three significant figures.

answer = time taken = 1099.24secs

Explanation:

Tthe detailed calculation is as shown in the attachment.

The principle applied is the resolution of resistance of circuit in series and in parralel and then finding the equivlaent resistance.

Use the ohm's law equation to find the current in the circuit.

Use the Power generated

From the specific heat capacity equation to get the time duration.

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Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

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2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

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