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11Alexandr11 [23.1K]
3 years ago
5

What would be an appropriate way to display the results of the experiment outlined?

Chemistry
2 answers:
miv72 [106K]3 years ago
6 0
The best way (in my opinion) is to put it in a table or spreadsheet. 
Elodia [21]3 years ago
3 0

Answer:

Use a table or spreasheet

Explanation:

This will depend of what kind of experiment you are doing, but in general terms, you can use a table or spreadsheet to display the results of any experiment.

Let's do an example, with a simple experiment. Let's suppose we are doing an experiment to determine the density of some objects, For this, we are going to use a rock, a ball and a little box.

The experiment states that in order to determine the density, you need to weight the objects, then calculate the volume, and finally calculate the density with the expression:

d = m/V

Now, let's suppose that you weight the 3 objects, and the values are:

rock: 150 g

ball: 300 g

box: 200 g

The volume for them are:

rock: 35 cm³

Ball: 45 cm³

Box: 64 cm³

Finally the density:

rock: 4.3 g/cm³

ball: 6.7 g/cm³

box: 3.1 g/cm³

These data, will be better understandable if you write a table like the one in picture 1. In that way, you show the data you have and the results of that experiment in a better way:

Download xlsx
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Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
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ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0
3 years ago
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