Answer:
Explanation:
Problem 1
<u>1. Data</u>
<u />
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
<u>2. Formula</u>
Since the temeperature is constant you can use Boyle's law for idial gases:

<u>3. Solution</u>
Solve, substitute and compute:


Problem 2
<u>1. Data</u>
<u />
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
<u>2. Formula</u>
You assume that the temperature does not change, and then can use Boyl'es law again.

<u>3. Solution</u>
This time, solve for V₂:

Substitute and compute:

You must round to 3 significant figures:

Problem 3
<u>1. Data</u>
<u />
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
<u>2. Formula</u>
At constant pressure, Charle's law states that volume and temperature are inversely related:

The temperatures must be in absolute scale.
<u />
<u>3. Solution</u>
a) Convert the temperatures to kelvins:
- T₁ = 25 + 273.15K = 298.15K
- T₂ = 35 + 273.15K = 308.15K
b) Substitute in the formula, solve for V₂, and compute:

You must round to two significant figures: 290 ml
Problem 4
<u>1. Data</u>
<u />
a) P = 865mmHg
b) Convert to atm
<u>2. Formula</u>
You must use a conversion factor.
Divide both sides by 760 mmHg

<u />
<u>3. Solution</u>
Multiply 865 mmHg by the conversion factor:

Answer:
they become water vapor and after that, they become a type of precipitation
For an object to sink in something, it's density has to be higher than the object it is in, so if it sinks in water the number has be higher than 1
And to float, and objects density has to be lower than that of the substance it is put in, so it has to be lower than 1.26
So between 1 and 1.26