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ra1l [238]
3 years ago
15

A circle with radius of \greenD{2\,\text{cm}}2cmstart color #1fab54, 2, start text, c, m, end text, end color #1fab54 sits insid

e a \blueD{11\,\text{cm} \times 12\,\text{cm}}11cm×12cmstart color #11accd, 11, start text, c, m, end text, times, 12, start text, c, m, end text, end color #11accd rectangle. What is the area of the shaded region? Round your final answer to the nearest hundredth.
Mathematics
1 answer:
HACTEHA [7]3 years ago
7 0

Answer:

119.43cm^2

Step-by-step explanation:

The computation of the area of the shaded region is shown below:

As we know that

= Area of the rectangle - Area of the circle

where,

Area of rectangle is

= length \times breadth

= 11\times 12

= 132 cm^2

And, the area of the circle is

=  \pi\times r^2\\\\ = \frac{22}{7} \times 2^2\\\\ = \frac{88}{7} cm^2

Now the area of the shaded region is

= 132 cm^2- \frac{88}{7}cm^2

= 119.43cm^2

We simply deduct the area of the circle from the area of the rectangle so that the area of the shaded region.

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Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodiu
Lesechka [4]

Answer:

Approximately 4.75.

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of  \rm CH_3COOH and \rm CH_3COO^{-} are equal.

{\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}

Since \rm CH_3COONa is a salt soluble in water. Once in water, it would readily ionize to give \rm CH_3COO^{-} and \rm Na^{+} ions.

Assume that the \rm CH_3COOH and \rm CH_3COO^{-} ions in this solution did not disintegrate at all. The solution would contain:

0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol of \rm CH_3COOH, and

0.06\; \rm mol of \rm CH_3COO^{-} from 0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol of \rm CH_3COONa.

Accordingly, the concentration of \rm CH_3COOH and \rm CH_3COO^{-} would be:

\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}.

\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}.

In other words, in this buffer solution, the initial concentration of the weak acid \rm CH_3COOH is the same as that of its conjugate base, \rm CH_3COO^{-}.

Hence, once in equilibrium, the \rm pH of this buffer solution would be the same as the {\rm pK}_{a} of \rm CH_3COOH.

Calculate the {\rm pK}_{a} of \rm CH_3COOH from its {\rm K}_{a}:

\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}.

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Answer:

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Step-by-step explanation:

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