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Shalnov [3]
4 years ago
8

Type the correct answer in the box. Jay, Kip, and Tia are each measuring the concentration of a solution. They are doing experim

ents separately. They are all are using a colorimeter—a device used for measuring colors. Jay repeats the experiment five times and takes the average measurement, while Kip performs the experiment once. Tia repeats the experiment three times but chooses one of the measurements instead of the average. The measurements taken by are least likely to contain random errors.
Chemistry
1 answer:
Kryger [21]4 years ago
3 0

Answer:

  • <em>The measurements taken by </em><em><u>Jay</u></em><em> are least likely to contain random errors.</em>

Explanation:

All experimental measures are subject to errors.

Even when the colorimeter is properly calibrated and correctly used, there are random errors.

Random errors are are due to fortuitous factors, such as minor oversight by the observer or small changes of the conditions under which the measurements are made.

You can minimize the random errors by increasing the number of measurements, because the random errors tend to happen in any direction; some measures will be greater and other will be less than the true value.

Chance will make that errors in on direction cancel with errors in the opposed direction, making the average the best measure.

Thus, <em>Jay</em>, by <em>repeating the experiment five times and taking the average measurement</em>, is making that<em> his measurements are</em> <em>least likely to contain random errors.</em>

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Using the following thermochemical data: 2Y(s) + 6HF(g) → 2YF3(s) + 3H2(g) ΔH° = –1811.0 kJ/mol 2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2
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ΔH° =   182.4 kJ/mol

Explanation:

The ΔH wanted is for the reaction :

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)

This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.

Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:

2YF3(s) + 3H2(g)  →  2Y(s) + 6HF(g)  ΔH° = 1811.0 kJ/mol (change sign)

dividing by two :

YF3(s) + 3/2H2(g)  →  Y(s) + 3HF(g)     ΔH° =  905.5 kJ/mol  Eq 1

2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol

dividing this one by two,

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

Now adding 1 and 2

YF3(s) + 3/2H2(g)  →  Y(s) + 3HF(g)     ΔH° =  905.5 kJ/mol  Eq 1

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

________________________________________________________

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g).   ΔH° =  905.5 + (-723.1) kJ/mol

ΔH° =   182.4 kJ/mol

Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.

8 0
4 years ago
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