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True [87]
3 years ago
15

Students are investigating the relationship between pressure and temperature in gases. They have blown up a balloon with carbon

dioxide from an air tank and
connected the balloon to a pressure gauge. The balloon is then submerged in ice or water baths at various temperatures in order to observe the pressure level
after several minutes. What is the flaw in the student's investigation, and how can the students improve their investigation to correctly determine the
relationship?
A The volume and density of the air in the balloon can change, therefore they need to use several different balloons at different volumes to account for
how density and volume affects pressure.
B. The volume of the balloon can change, therefore they need to use a rigid container such as plastic or metal to hold the air
C. The volume of the balloon can change, therefore they need to use several different balloons at different volumes to account for how volume atteots
pressure.
D. The volume and density of the air in the balloon can change, therefore they need to use a rigid container such as plastic or metal to hold the air
Chemistry
2 answers:
Ierofanga [76]3 years ago
8 0

Answer:

b

Explanation:

Alik [6]3 years ago
6 0
B: the volume of the balloon can change. Therefore they need to use several different balloons at different volumes to account for how volume atteots
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A sample of gas occupies 280 mL when the pressure is 560.00 mm Hg . If the temperature remains constant , what is the new pressu
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Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 560.00 mmHg

P_2 = final pressure = ?

V_1 = initial volume = 280 mL

V_2 = final volume = 560.0 mL

Now put all the given values in the above formula, we get:

560.00mmHg\times 280 mL=P_2\times 560.0 mL

P_2=280mmHg

Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg

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