experiment
a pair of 100 ml samples of water are taken from a well bored into a large underground salt (nacl) deposit. sample #1 is from the top of the well, and is initially at 32 °c. sample #2 is from a depth of 50. m, and is initially at 42 °c. both samples are allowed to come to room temperature (20. °c) and 1 atm pressure. an nacl precipitate is seen to form in sample #1.
a student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drink. the student opens the bottle, and hears a loud hiss as gas under pressure escapes from the bottle.
(predicted observation (choose one))
a bigger mass of nacl precipitate will form in sample #2.
a smaller mass of nacl precipitate will form in sample #2.
the same mass of nacl precipitate will form in sample #2.
no precipitate will form in sample #2.
i need more information to predict whether and how much precipitate will form in sample #2.
the bubbles will shrink, and some may vanish.
the bubbles will grow, and more may appear.
the bubbles wont change.
i need more information to predict what will happen to the bubbles.

Question

Accepted Answer

### First Experiment (Water Samples and NaCl Precipitation)
# Explanation:
## Step1: Analyze Solubility vs Temperature/Pressure
Solubility of NaCl in water generally increases with temperature, but here Sample #1 is from the top (32°C, surface pressure ~1 atm), Sample #2 is from 50 m depth (initially 42°C, but when brought to 20°C and 1 atm, we need to consider solubility changes. However, the key is that at lower temperature (20°C), solubility of NaCl is lower than at higher temps (32°C, 42°C). But Sample #2 was at 42°C initially (higher temp, higher solubility) and is now cooled to 20°C. Wait, no—Sample #1 is initially 32°C (top, 1 atm), Sample #2 is initially 42°C (50 m, pressure higher than 1 atm). When both are brought to 20°C and 1 atm, we need to see which has more dissolved NaCl initially. But the problem states "a pair of 100 mL samples of water are taken from a well bored into a large underground salt (NaCl) deposit." So the water is saturated or near-saturated with NaCl. When cooled, solubility decreases, so precipitation occurs. But Sample #2 was at higher temperature (42°C) initially, which has higher solubility than 32°C (Sample #1). So when both are cooled to 20°C, the one that was at higher temp (Sample #2) had more dissolved NaCl initially (since higher temp = higher solubility for NaCl in water). Wait, no—solubility of NaCl in water increases slightly with temperature. So at 42°C, solubility is higher than at 32°C, which is higher than at 20°C. So Sample #2 (initially 42°C) has more dissolved NaCl than Sample #1 (initially 32°C) when taken. Then, when both are cooled to 20°C (lower solubility than both 32 and 42), the amount of excess NaCl (to precipitate) depends on initial dissolved amount. Since Sample #2 had more dissolved (because higher initial temp, higher solubility), when cooled to 20°C, the amount of NaCl that will precipitate should be more? Wait, no—wait, the initial conditions: Sample #1 is from top (32°C, 1 atm), Sample #2 from 50 m (42°C, higher pressure). When brought to 20°C and 1 atm, the pressure drops for Sample #2 (from >1 atm to 1 atm), but for NaCl, pressure effect on solubility is minimal. Temperature is the main factor. So solubility at 20°C is S20, at 32°C is S32, at 42°C is S42, with S42 > S32 > S20. So Sample #1 has dissolved NaCl = S32 (saturated at 32°C), Sample #2 has dissolved NaCl = S42 (saturated at 42°C). When cooled to 20°C, the maximum dissolved NaCl is S20. So the amount of precipitate in Sample #1 is S32 - S20, in Sample #2 is S42 - S20. Since S42 > S32, then S42 - S20 > S32 - S20, so more precipitate in Sample #2? Wait, but the first experiment's predicted observation options: "A bigger mass of NaCl precipitate will form in Sample #2" – but wait, the original experiment says "An NaCl precipitate is seen to form in Sample #1". Wait, maybe I got it reversed. Wait, Sample #1 is initially 32°C (top, 1 atm), Sample #2 is initially 42°C (50 m, higher pressure). When brought to 20°C and 1 atm, Sample #1 was at 32°C (solubility S32), now at 20°C (S20 < S32), so precipitate. Sample #2 was at 42°C (S42 > S32 > S20), so when cooled to 20°C, the excess is S42 - S20, which is more than S32 - S20 (Sample #1's excess). But the option is "A bigger mass of NaCl precipitate will form in Sample #2" – but wait, maybe the initial temperature of Sample #2 is 42°C, but when brought to 20°C, the solubility drops, but maybe the pressure change (from 50 m depth, which is ~5 atm, to 1 atm) – but for NaCl, pressure has little effect. So the main factor is temperature. So if Sample #2 was at higher temp, more dissolved, so more precipitate when cooled. But the first experiment's correct prediction? Wait, maybe the answer is "The same mass of NaCl precipitate will form in Sample #2" – no, that doesn't make sense. Wait, maybe I made a mistake. Alternatively, maybe the solubility of NaCl in water is relatively insensitive to temperature, so the amount of precipitate is similar. But the options include "I need more information to predict whether and how much precipitate will form in Sample #2" – but the first experiment's predicted observation: the correct option? Wait, the first experiment's problem: two 100 mL samples, Sample #1 (top, 32°C, 1 atm) has precipitate at 20°C, 1 atm. Sample #2 (50 m, 42°C, initially) is brought to 20°C, 1 atm. To predict, we need to know solubility at 32, 42, 20°C. But since we don't have exact solubility values, the correct prediction is "I need more information to predict whether and how much precipitate will form in Sample #2". Wait, but the options for the first experiment: the starred option? Wait, the image shows the first experiment's predicted observation options, with a star on "A bigger mass of NaCl precipitate will form in Sample #2" – but maybe that's incorrect. Alternatively, maybe the key is that Sample #2 was at higher temperature (42°C) which has higher solubility, so when cooled to 20°C (lower solubility), more NaCl will precipitate than Sample #1 (which was at 32°C, lower initial solubility). So "A bigger mass of NaCl precipitate will form in Sample #2" is correct.

## Step2: Analyze the Soft Drink Experiment
For the soft drink (carbonated, CO₂ dissolved under pressure), when the bottle is opened, pressure drops to 1 atm (from higher pressure inside). The solubility of CO₂ in water is pressure-dependent (Henry's Law: solubility ∝ pressure). So when pressure decreases, solubility of CO₂ decreases, so the dissolved CO₂ comes out as gas bubbles. Also, heating (when the student heats the bottle? Wait, the student opens the bottle and hears a hiss – opening releases pressure, causing bubbles. But the predicted observation: the bubbles are tiny initially, then when opened, pressure drops, so solubility of CO₂ decreases, so more gas comes out, bubbles grow and more appear. So the correct option is "The bubbles will grow, and more may appear."

# Answer:
### First Experiment (NaCl Precipitation)
Answer: A bigger mass of NaCl precipitate will form in Sample #2.
(Explanation: Solubility of NaCl increases with temperature. Sample #2 was initially at 42°C (higher temp, more dissolved NaCl) than Sample #1 (32°C). When both are cooled to 20°C (lower solubility), the excess NaCl (to precipitate) is greater for Sample #2, so a bigger mass precipitates.)

### Second Experiment (Soft Drink Bubbles)
Answer: The bubbles will grow, and more may appear.
(Explanation: Opening the bottle reduces pressure, decreasing CO₂ solubility (Henry’s Law). Dissolved CO₂ forms gas bubbles, which grow as more gas escapes, and more bubbles appear as solubility drops further.)

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QUESTION IMAGE

Question

Explanation:

First Experiment (Water Samples and NaCl Precipitation)

Step1: Analyze Solubility vs Temperature/Pressure

Answer:

First Experiment (NaCl Precipitation)