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 3.00 atmospheres. What would the volume of the balloon up there? (Temperature is constant)

0.62 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 785 mmHg

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

Step 2:

Conversion of the pressure in mmHg to atm.

It is important to express the initial and the final pressure in the same unit. Either express both in atm or in mmHg. What ever the case is, we'll still arrive at same answer. Here, we shall be converting from mmHg to atm. This is illustrated below:

760mmHg = 1atm

Therefore, 785 mmHg = 785/760 = 1.03 atm

Step 3:

Determination of the final volume. This is illustrated below.

We shall be applying the Boyle's law equation since the temperature is constant.

P1V1 = P2V2

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 1.03 atm

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

P1V1 = P2V2

1.03 x 1.8 = 3 x V2

Divide both side by 3

V2 = (1.03 x 1.8) /3

V2 = 0.62 L

Therefore, the new volume of the balloon is 0.62 L

there all right good

explanation:

it's c. elements are placed in rows based on how may valencies electrons have.