Question

use the following information to answer question 8.
snowshoes allow users to walk on soft snow without sinking.
a student takes a step with a snowshoe, exerting 600 n of force on the ground. the snowshoe has an area of 0.15 m².

1. when taking a step, the student exerts a pressure of ________ on the ground.

a. 0.025 n/m²
b. 90 n/m²
c. 615 n/m²
d. 4000 n/m²

use the following information to answer question 9.
the following graph shows how many grams of baking soda will dissolve into 250 ml of water at various temperatures.
solubility of baking soda in 250 ml of water

1. a student fills a beaker with 100 ml of water at 25 °c. the concentration of the baking soda solution in the beaker is

a. 1 g/100 ml
b. 5 g/100 ml
c. 10 g/100 ml
d. 25 g/100 ml

Explanation:

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<reasoning>

Calculate pressure exerted on the ground

Using the Pressure Calculation knowledge point
\[

$$\begin{aligned} F &= 600\text{ N}\\ A &= 0.15\text{ m}^2\\ P &= \frac{F}{A} = \frac{600}{0.15} = 4000\text{ N/m}^2 \end{aligned}$$

\]

Read solubility from the graph

Using the Solubility Curves knowledge point
\[

$$\begin{aligned} T &= 25^\circ\text{C}\\ \text{Mass in } 250\text{ mL} &\approx 25\text{ g} \end{aligned}$$

\]

Calculate concentration of the solution

Using the Solution Concentration knowledge point
Since the student is at a "Mastered" level for this concept, we keep this step concise.
\[

$$\begin{aligned} \text{Concentration} &= \frac{25\text{ g}}{250\text{ mL}} = 0.1\text{ g/mL}\\ \text{Concentration per } 100\text{ mL} &= 0.1\text{ g/mL} \times 100\text{ mL} = 10\text{ g/100 mL} \end{aligned}$$

\]
</reasoning>

<answer>

Question 8

<mcq-option>A. \(0.025\text{ N/m}^2\)</mcq-option>
<mcq-option>B. \(90\text{ N/m}^2\)</mcq-option>
<mcq-option>C. \(615\text{ N/m}^2\)</mcq-option>
<mcq-correct>D. \(4000\text{ N/m}^2\)</mcq-correct>

Question 9

<mcq-option>A. \(1\text{ g/100 mL}\)</mcq-option>
<mcq-option>B. \(5\text{ g/100 mL}\)</mcq-option>
<mcq-correct>C. \(10\text{ g/100 mL}\)</mcq-correct>
<mcq-option>D. \(25\text{ g/100 mL}\)</mcq-option>
</answer>

<post_analysis>
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"question_type": "Multiple Choice",
"knowledge_point": [
"Natural Science",
"Physics",
"Pressure Calculation"
]
}
</post_analysis>

Answer:

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"concepts_used": [
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</pre_analysis>

<reasoning>

Calculate pressure exerted on the ground

Using the Pressure Calculation knowledge point
\[

$$\begin{aligned} F &= 600\text{ N}\\ A &= 0.15\text{ m}^2\\ P &= \frac{F}{A} = \frac{600}{0.15} = 4000\text{ N/m}^2 \end{aligned}$$

\]

Read solubility from the graph

Using the Solubility Curves knowledge point
\[

$$\begin{aligned} T &= 25^\circ\text{C}\\ \text{Mass in } 250\text{ mL} &\approx 25\text{ g} \end{aligned}$$

\]

Calculate concentration of the solution

Using the Solution Concentration knowledge point
Since the student is at a "Mastered" level for this concept, we keep this step concise.
\[

$$\begin{aligned} \text{Concentration} &= \frac{25\text{ g}}{250\text{ mL}} = 0.1\text{ g/mL}\\ \text{Concentration per } 100\text{ mL} &= 0.1\text{ g/mL} \times 100\text{ mL} = 10\text{ g/100 mL} \end{aligned}$$

\]
</reasoning>

<answer>

Question 8

<mcq-option>A. \(0.025\text{ N/m}^2\)</mcq-option>
<mcq-option>B. \(90\text{ N/m}^2\)</mcq-option>
<mcq-option>C. \(615\text{ N/m}^2\)</mcq-option>
<mcq-correct>D. \(4000\text{ N/m}^2\)</mcq-correct>

Question 9

<mcq-option>A. \(1\text{ g/100 mL}\)</mcq-option>
<mcq-option>B. \(5\text{ g/100 mL}\)</mcq-option>
<mcq-correct>C. \(10\text{ g/100 mL}\)</mcq-correct>
<mcq-option>D. \(25\text{ g/100 mL}\)</mcq-option>
</answer>

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"question_type": "Multiple Choice",
"knowledge_point": [
"Natural Science",
"Physics",
"Pressure Calculation"
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