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evablogger [386]

2 years ago

9

in training to run a half marathon, jenny ran 2/5 hours on tuesday, 1 1/6 hours on thursday, and 2 1/15 on Saturday. what is the

total amount of hours that jenny ran this week? (simplify your answer and state it as a MIXED NUMBER)

2 answers:

Xelga [282]2 years ago

7 0

9514 1404 393

Answer:

3 19/30 hours

Step-by-step explanation:

The given fractions will have 30 as a common denominator.

12/30 + 1 5/30 + 2 2/30 = (1 +2) +(12 +5 +2)/30 = 3 19/30 . . . hours

Jenny ran 3 19/30 hours this week.

tamaranim1 [39]2 years ago

7 0

Answer:

Step-by-step explanation:

Total hours that Jenny ran = 3.63 hours.

Step-by-step explanation:

Jenny ran on Tuesday for = 2/5 hours or 0.4 hours.

Time consumed to run on Thursday = 11/6 hours or 1.83 hours.

Time consumed to run on Saturday = 21/ 15 hours or 1.4 hours.

Here, the total hours can be calculated by just adding all the running hours. So the running hours of Tuesday, Thursday, and Saturday will be added to find the total hours.

Total hours that Jenny ran = 0.4 + 1.83 + 1.4 = 3.63 hours.

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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.

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$0.06\; \rm mol$ 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:

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$\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

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