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Westkost [7]
3 years ago
12

In all measurements, you are looking for values that are accurate. Pretend that a pebble with a known mass of 0.567 g was massed

four times where the following masses were obtained: 0.256 g, 0.723 g, 0.554 g and 0.354 g. Those masses are an example of:__________.
a) being precise but not accurate.
b) being both precise and accurate.
c) being neither precise nor accurate.
d) being accurate but not precise.
Chemistry
1 answer:
givi [52]3 years ago
4 0

Answer:

The correct answer is option c, that is, being neither precise nor accurate.

Explanation:

Obtaining the experimental values that come almost close to the true value is termed as accuracy. On the other hand, precision is obtaining experimental values continuously, which may come either far away or near to the true value.  

Based on the given information, it is clear that 0.554 grams, that is, the third value obtained from the experiment is showing some closeness to the true value, which is 0.567 grams. While, neither any of the other values are close or is coming near to the true value. This shows that the measurements are an illustration of being neither precise nor accurate.  

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Answer:

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Explanation:

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\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

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\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

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\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

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\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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