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solmaris [256]
2 years ago
8

Base your answer on the diagram below and on your knowledge of Earth Science. The diagram represents a cutaway view of a flat-pl

ate solar collector used to heat water at a New York State location. Explain why the flow tubes and collector plate inside the solar collector are black in color

Chemistry
1 answer:
algol132 years ago
3 0

Answer:

Explanation:

A solar collector is a device that absords Sun's heat energy to heat air or water. It is majorly used for heating purpose, and do not generate electricity directly.

The flow tubes and collector plate are black in color so as to increase the intensity of heat generated by the collector. A black body is a good absorber of heat, it absorbs most heat directed to the collector. Also, a black body is a good radiator of heat, the heat absorbed is rediated to the appropriate channels for the heating of water or air molecules. The black color increases the efficeincy and percentage of solar energy absorbed by the collector.

If a reflective color is used (e.g white), major percentage of the incident heat would be reflected. This would decrease the efficiency of the solar collector.

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The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.
olchik [2.2K]

Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
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Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

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\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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