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Naya [18.7K]
4 years ago
9

Give examples from our daily lives on each of the 4 factors that affect the rate of the reaction. (Concentration, Surface Area,

Temperature, Catalysts)
Chemistry
1 answer:
LenKa [72]4 years ago
6 0
Chemists use reactions to generate a desired product. For the most part, a reaction is only useful if it occurs at a reasonable rate. For example, a reaction that took 8,000 years to complete would not be a desirable way to produce brake fluid. However, a reaction that proceeded so quickly that it caused an explosion would also not be useful (unless the explosion was the desired result). For these reasons, chemists wish to be able to control reaction rates. In order to gain this control, we must first know what factors affect the rate of a reaction. We will discuss some of these factors in this section.
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A chemist requires 5.00 liters of 0.420 M H2SO4 solution. How many grams of H2SO4 should the chemist dissolve in water? 129 gram
natali 33 [55]

Answer:

Approximately 206 grams.

Explanation:

How many moles of sulfuric acid \mathrm{H_2SO_4} are there in this solution?

\text{Number of moles of solute} = \text{Concentration} \times \text{Volume}.

The unit for concentration "\mathrm{M}" is equivalent to mole per liter. In other words, \rm 1\;M = 1\; mol\cdot L^{-1}. For this solution, the concentration of \mathrm{H_2SO_4} is \rm 0.420\;M = 0.420\; mol\cdot L^{-1}.

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V\\&= \rm 0.420\;mol\cdot L^{-1}\times 5.00\; L \\&= \rm 2.10\; mol\end{aligned}.

What's the mass of that \rm 2.10\; mol of \mathrm{H_2SO_4}?

Start by finding the molar mass M of \mathrm{H_2SO_4}.

Relative atomic mass data from a modern periodic table:

  • H: 1.008;
  • S: 32.06;
  • O: 15.999.

\displaystyle M(\mathrm{H_2SO_4}) = 2\times \underbrace{1.008}_{\mathrm{H}} + 1\times \underbrace{32.06}_{\mathrm{S}} + 4\times \underbrace{15.999}_{\mathrm{O}} = \rm 98.072\;g\cdot mol^{-1}.

\text{Mass} = \text{Quantity in moles} \times \text{Molar Mass}.

m = n \cdot M = \rm 2.10\; mol \times 98.072\;g\cdot mol^{-1} \approx 206\; g.

In other words, the chemist shall need approximately 206 grams of \mathrm{H_2SO_4} to make this solution. As a side note, keep in mind that the 206 grams of \mathrm{H_2SO_4} also take up considerable amount of space. Therefore it will take less than 5.00 L of water to make the 5.00 L solution.

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