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Finger [1]
3 years ago
6

A sample of an element has a mass of 34.261 grams and a volume of 3.8 cubic centimeters. To which number of significant figures

should the calculated density of the sample be expressed?
(1) 5 (3) 3
(2) 2 (4) 4
Chemistry
1 answer:
nikklg [1K]3 years ago
3 0
The answer is (2). To calculate the density, you need to divide the mass with volume. When doing division, the significant figures of result is the minimal of the numbers using before. So the answer is 2.
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25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
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Answer:

Concentration of the original \rm KOH solution: approximately 1.05\; \rm mol \cdot dm^{-3}.

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\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}.

\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 0.0350\; \rm dm^{3} of 0.75\; \rm mol \cdot dm^{3} \rm HCl\! solution:

\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}.

\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

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Question 1 (1 point)
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