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2 years ago

Help someone to help me :( in number 2 and 3

Mathematics

1 answer:

sveticcg [70]2 years ago

7 0

So the things that you need to know are in the picture I have to see if I can help you with the other one hope this helps.

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In Ms. Henrick's class, there are 10 boys and 12 girls. What is the ratio of boys to girls? Express your answer as a fraction in

vodomira [7]

Answer:

the ratio is 10:12 and if you were to write it in a fraction you get 10/12 which can be simplified as 5/6

5 0

2 years ago

Read 2 more answers

What is the measurement of ∠BAC and ∠ABC? Explain your process of solving.

Paraphin [41]

Answer:

<BAC = 78

<ABC = 68

Step-by-step explanation:

The remote angles theorem states that when one extends a side of a triangle, the angle formed between the extension and one of the sides of the triangle is equal to the sum of the two non-adjacent angles inside the triangle. One can apply this theorem here and state the following,

<BAC + <ABC = <ACD

Substitute,

(5y + 3) + (4y + 8) = (146)

Simplify,

9y + 11 = 146

Inverse operations,

9y + 11 = 146

-11 -11

9y = 135

/9 /9

y = 15

Now substitute this value back into the expressions to find the numerical measurement of (<BAC) and (<ABC),

<BAC = 5y + 3

5(15) + 3

<ABC = 4y + 8

4(15) + 8

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

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