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professor190 [17]
2 years ago
6

Solve by factoring2n^2+12n+18

Mathematics
1 answer:
Gelneren [198K]2 years ago
8 0

Answer:

Step-by-step explanation:

d/dn(2 n^2 + 12 n + 18) = 4 (n + 3)

Indefinite integral:

integral(18 + 12 n + 2 n^2) dn = (2 n^3)/3 + 6 n^2 + 18 n + constant

Global minimum:

min{2 n^2 + 12 n + 18} = 0 at n = -3

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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}.

7 0
3 years ago
Factorise: (Using Algebraic Identities)<br> 64x^3-(4x-3y)^3
Naddika [18.5K]

Answer:

9y(16x2−12xy+3y2)

Step-by-step explanation:

Factor 64x3−(4x−3y)3

144x2y−108xy2+27y3

=9y(16x2−12xy+3y2)

Answer:

9y(16x2−12xy+3y2)

6 0
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