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Elena-2011 [213]
3 years ago
11

You're at the zoo and have a big red 1.80 L helium balloon. The barometric pressure today is 785 mmHg. Then you hear the roar of

a lion. Startled, you accidentally release the balloon. It flies away. By the time it reaches the clouds, the atmospheric pressure that high is only 0.300 atmospheres. What would be the volume of the balloon up there?
Chemistry
1 answer:
fomenos3 years ago
8 0

Answer: The volume of the balloon up there is 6.192 L.

Explanation:

Given: V_{1} = 1.80 L,    P_{1} = 785 mm Hg (mm Hg = 0.00131579) = 1.032 atm

P_{2} = 0.300 atm,       V_{2} = ?

Formula used to calculate volume is as follows.

P_{1}V_{1} = P_{2}V_{2}

Substitute the value into above formula as follows.

P_{1}V_{1} = P_{2}V_{2}\\1.032 atm \times 1.80 L = 0.300 atm \times V_{2}\\V_{2} = 6.192 L

Thus, we can conclude that the volume of the balloon up there is 6.192 L.

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A certain electrolyte solution contains 1 gram of salt for every 8 grams of sugar and every 200 grams of water. If the sugar to
12345 [234]

Answer:

The resulting solution contains approximately 666 g of water.

Explanation:

In the initial solution we have:

1g salt : 8g sugar : 200g water

This means that the ratios are:

\frac{salt}{sugar}  = \frac{1}{8} \\\\\frac{sugar}{water} = \frac{8}{200} =\frac{1}{25}

In the final solution we have:

5g salt: xg sugar: yg water

The new ratios are:

\frac{salt}{sugar} = \frac{3}{8} \\\\\frac{sugar}{water} = \frac{1}{50}

Now we can calculate the amount of sugar in the final solution:

\frac{salt}{sugar}  = \frac{5}{x} =\frac{3}{8} \\\\X = 13.333 g

Finally, we calculate the amount of water:

\frac{sugar}{water} = \frac{13.333}{y} = \frac{1}{50} \\y = 666.667 g

4 0
3 years ago
An exhaled air bubble underwater at 290.
Vinil7 [7]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 290 kPa

Final pressure (P_{2}) = 104 kPa

Initial volume (V_{1}) = 18.9 ml

To find:

Final volume (V_{2})

We know;

From the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

   P × V = constant

   P ∝ \frac{1}{V}

From the above equation;

              \frac{P_{1} }{P_{2} }  = \frac{V_{2} }{V_{1} }

P_{1} represents the initial pressure of the gas

P_{2} represents the final pressure of the gas

V_{1} represents the initial volume of the gas

V_{2} represents the final volume of the gas

Substituting the values of the above equation;

                    \frac{290}{104} = \frac{V_{2} }{18.9}

             V_{2} = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

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