Answer 1) : The density of the hot air inside the balloon can be found out by using ideal gas equation;
PV = nRT;
As n is number of moles and in gases, number of moles along with mass per mole is equal to the density of the gas.
If the moles in the gas are more the density will be more.
here, density (ρ) = mass (m) / volume (V); substituting in the ideal gas equation we get,
ρ = mP / RT
Answer 2) ρ (hot air) = ρ (cold air) X 
Here according to the formula because T(hot air) >T(cold air),
So, the density of hot air greater than the density of cold air.
The relationship between the ρ (h) = ρ(c) X
Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
The location of the valence electron or the outermost electron is expressed in quantum numbers. There are five quantum numbers: prinicipal (n), angular momentum (l), magnetic (ms) and magnetic spin (ms) quantum numbers. This is based on Bohr's atomic model where electrons orbit around the nucleus. These electrons are in the orbitals with specific energy levels. Starting from energy level 1 that is closest to the nucleus, the energy level decreases to 2, 3, 4, 5, 6, and 7. These energy level numbers represent the principal quantum number. Within each orbital also contains subshell. From increasing to decreasing order, these subshells are the s, p, d and f subshells. These subshells represent the angular momentum quantum numer. Specifically, s=0, p=1, d=2 and f=3. Therefore, if the electron is in the orbital 5p, the quantum number would be: 5, 1. Applying these to the choices, the correct pairing would be:
2p: n=2. l=1
3d: n=3, l=2
2s: n=2. l=0
4f: n=4. l=3
1s: n=1, l=0
<u>Answer:</u>
<u>For a:</u> Lead iodide is a yellow precipitate.
<u>For b:</u> Barium sulfate is a white precipitate.
<u>For c:</u> Ferric hydroxide is a brown precipitate.
<u>For d:</u> Copper (II) hydroxide is a blue precipitate.
<u>Explanation:</u>
Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.
For the given options:
The chemical reaction between KI and lead (II) nitrate follows:
The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.
The chemical reaction between barium chloride and sulfuric acid follows:
The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.
The chemical reaction between NaOH and ferric chloride follows:
The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.
The chemical reaction between NaOH and copper sulfate follows:
The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.
