All categories

2 years ago

Simplify 6(2+4x)+3(x2-5x)how do i do this?

Mathematics

1 answer:

babunello [35]2 years ago

3 0

Step 1: Distribute

So it becomes 12+24x+6x-15x

Step 2: Combine like-terms!

Then it becomes 15x+12 (It's 15x because 24x+6x= 30x then "add" the last -15x.
(30+ -15 = 15)

You might be interested in

Suppose that 1 month before the election a random sample of 500 registered voters are surveyed. From this sample 270 indicate th

Aleonysh [2.5K]

Answer:

[ 0.4964, 0.5836 ]

Step-by-step explanation:

Data provided in the question:

Total sample size = 500

person voting for smith = 270

thus,

P( person voting for smith ), p = $\frac{270}{500}$ = 0.54

Confidence level = 95%

now,

standard error, SE = $\sqrt{\frac{p(1-p)}{n}$

SE = $\sqrt{\frac{0.54(1-0.54)}{500}$

SE = 0.0223

now,

Confidence interval = p ± ( z × SE )

here,

z value for 95% confidence interval is 1.96

Confidence interval = [ 0.54 - ( 1.96 × 0.0223 ), 0.54 + ( 1.96 × 0.0223 ) ]

= [ 0.54 - 0.0436 , 0.54 + 0.0436 ]

= [ 0.4964, 0.5836 ]

8 0

2 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 ${\rm K}_{a}$ :

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