Explanation:
1) Initial mass of the Cesium-137=
= 180 mg
Mass of Cesium after time t = N
Formula used :
Half life of the cesium-137 =
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant

Now put all the given values in this formula, we get
Mass that remains after t years.

Therefore, the parent isotope remain after one half life will be, 100 grams.
2)
t = 70 years


N = 35.73 mg
35.73 mg of cesium-137 will remain after 70 years.
3)


N = 1 mg
t = ?

t = 224.80 years ≈ 225 years
After 225 years only 1 mg of cesium-137 will remain.
Answer: The new volume of the gas is smaller.
Explanation:
Volume and pressure are inversely proportional (as one goes up, one goes down). So as you increase the pressure, you decrease the volume.
2 elections will fill the first energy level
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:

Explanation:
Hello there!
In this case, since the vaporization process is carried out in order to turn a liquid into a gas due to the addition of heat, we can use the following heat equation involving the heat of vaporization of water or any other substance:

Thus, since this heat of vaporization for water is 2259.36 J/g, we plug in this amount to obtain the total energy for this process.

Which is positive due to the necessity of heat.
Regards!