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olya-2409 [2.1K]
3 years ago
5

What is the simplified form of 24r - 3(2r - 5t) - 4t?

Mathematics
2 answers:
Drupady [299]3 years ago
6 0
<span>24r - 3(2r - 5t) - 4t
= 24r - 6r + 15t - 4t
= 18r + 11t

answer
</span><span>18r + 11t</span>
inysia [295]3 years ago
4 0
-3(2r-5t)=-6r+15t
24r-6r+15t-4t=
18r+11t is the simplified form
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Look at mina's work below 10 7/12 - 9/12 = (10/12 + 7/12) - 9/12 = 17/12 - 9/12 = 8/12 is mina's solution reasonable. Explain
RSB [31]
10 7/12 - 9/12=(10/12+7/12)-9/12=17/12-9/12=8/12

127/12-9/12

Nope. Her solution doesn't seem reasonable. In order to subtract 9/12 from 10 7/12 you must convert 10 7/12 to an improper fraction. You should then have 127/12 where she has 10/12.

Hope this helps! =)



8 0
3 years ago
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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that is negative, as we already said.

Then we have:

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