Assuming the balloon initially has volume of 0 when deflated, the total P(deltaV) = (1.00 atm)(550,000 ft^3) = 550,000 atm-ft^3. To convert into work units, we can first convert ft^3 to L:
(550,000 atm-ft^3)(1 m/3.28 ft)^3
= (15,586.2 atm-m^3)
Then we convert to L:
(15,586.2 atm-m^3)(1000 L/m^3)
= 15,586,200 atm-L
Then we convert to J:
(15,586,200 atm-L)(101.325 J / 1 atm-L)
= 1.579 x 10^9 J
Answer: I think it would be false but im not 1000% sure
Answer:
The object will be sank
Explanation:
In this case the object is more dense than water.
Density is the relationship between a certain amount of mass of matter and the volume that is being occupied by it.
The object occupies more volume, so it occupies more mass.
As the mass from the object is higher, the object will be sank because the weight is higher than the weight from the liquid.
If the object has a lower density than the water, it will float on it.
PH = -log [H3O+]
4.15 = -log [H3O+]
[H3O+] = 10^(-4.15)
[H3O+]= 7.08 × 10^-5
