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topjm [15]
2 years ago
10

A hiker seals a plastic bag that contains dried fruits before he climbs a mountain. What will most likely happen to the plastic

bag as the hiker reaches the top of the mountain?
The bag will expand.
The bag will shrink.
The bag will remain the same size.
The bag will allow air to enter
Chemistry
2 answers:
Sonja [21]2 years ago
4 0
I think the correct answer would be the first option. As the hiker reaches the top of the mountain, the plastic bag would, most likely, expand. Before starting, the plastic bag would have a certain amount of gas particles inside. As the you hike, the pressure would decrease causing for the gas molecules to reestablish equilibrium thus pushing the bag outwards. In other words, expanding the bag. Also, base from the ideal gas equation, PV = nRT, at constant temperature and number of moles, pressure and volume are inversely related. Thus, as the pressure decreases, the volume of the system would increase accordingly.
defon2 years ago
3 0

Answer: A

Explanation:

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2 years ago
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Thermal energy would naturally flow between objects at which temperatures?
marissa [1.9K]

Answer:

i think the answer is letter C. From 35°c to 45°c

Explanation:

sorry if it is wrong

4 0
2 years ago
At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements
Gnom [1K]

Given that:

  • At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

     Initial volume of    Final volume of    Mass of water   Density (g/mL)

     burette (mL)        burette   (mL)       dispensed (g)

 

Sample 1      2.33                     7.34                   5.000               -----

Sample 2      7.34                    12.37                 5.025                -----

Sample 3      12.37                   18.50                6.112                  -----

Sample 4      18.50                  24.57               6.064                 -----

Sample 5     24.57                  31.31                6.720                  -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

  • The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

\mathbf{density = \dfrac{mass}{volume}}

\mathbf{density = \dfrac{5.01 g}{5.000 ml}}

density = 0.998004 g/ml

Sample 2:

\mathbf{density = \dfrac{5.025 g}{5.03ml}}

density = 0.999006 g/ml

Sample 3:

\mathbf{density = \dfrac{6.112 g}{6.13ml}}

density = 0.997064 g/ml

Sample 4:

\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}

density = 0.999012 g/ml

Sample 5:

\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}

density = 0.997033 g/ml

Thus, the average density for all the samples is:

\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 +   0.999012  + 0.997033  )}{5}}

= 0.998024

∴

The percentage error for the two densities measurement is:

=\dfrac{ (experimental \  value -theoretical  \ value)\times 100 }{theoretical  \ value}

Given that the theoretical value = 0.997044 g/ml

Then;

\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0
2 years ago
An automobile tire contains air at 320.×103 Pa at 20.0 ◦C. The stem valve is removed and the air is allowed to expand adiabatica
NISA [10]

Answer:

6.15.3 k

Explanation:

From the question we can see that

q = 0,  Δu = w

Then,

T_f = \frac{C_{V,m}+RP_{ext}P_i}{C_{V,m}+RP_{ext}P_f} T_i

putting values wet

=\frac{2.5\times 8.314+8.314\left(10^5\right)\left(3.20\times 10^5\right)}{2.5\times 8.314+\left(8.314\right)\left(10^5\right)\left(10^5\right)}\times \:293

T_f = 615.3 K

6 0
3 years ago
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