Weight of the balloon = 2.0 g
Six weights each of mass 30.0 g is added to the balloon.
Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g
Density of salt water = 1.02 g/mL
Calculating the volume from mass and density:

Converting the volume from mL to cubic cm:

Assuming the balloon to be a sphere,
Volume of the sphere =
π

r = 3.49 cm
Radius of the balloon = 3.49 cm
Diameter of the balloon = 2 r = 2*3.49 cm = 6.98cm
Answer:
The question you have provided is incomplete
Explanation:
Answer: P₂=0.44 atm
Explanation:
For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.
Gay-Lussac's Law: 
First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.
64+273K=337K
Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.


P₂=0.44 atm
Answer:
The minimum concentration of Cl⁻ that produces precipitation is 12.6M
Explanation:
The Ksp of PbCl₂ is expressed as:
PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)
The Ksp is:
Ksp = 1.6 = [Pb²⁺] [Cl⁻]²
When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.
A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:
1.6 = [0.010M] [Cl⁻]²
160 = [Cl⁻]²
12.6M = [Cl⁻]
<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>