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Novay_Z [31]
3 years ago
14

Simplify: {[(16 ÷ 4) × (2 × 6)] ÷ 6} + 4 =

Mathematics
2 answers:
Mashcka [7]3 years ago
5 0
[4 x 12 divide 6] + 4
[48 divide 6] + 4
[8]+ 4
12
Trava [24]3 years ago
3 0
<span>{[(16 ÷ 4) × (2 × 6)] ÷ 6} + 4 = 
</span><span>{[(4) × (12)] ÷ 6} + 4 = 
</span>{[4 × 12] ÷ 6} + 4 = 
{[48] ÷ 6} + 4 = 
{48 ÷ 6} + 4 = 
{8} + 4 = 
8 + 4 = 
12
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1^7 is the answer you are looking for

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12

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( PLASE HELP ME) Solve the following equation.
vovikov84 [41]

Answer:

x = 2

Step-by-step explanation:

5(x-4) =-10

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3 years ago
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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
4x - 17 = 32 what is x
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X = 12.25 and I’m typing this to get to 20 characters
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