<u>Answer:</u>
P2 = 778.05 mm Hg = 1.02 atm
<u>Explanation:</u>
We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.
For this, we would use the equation:
where P1 = 724 mm hg
V1 = 3.05 L
T1 = 298 K
P2 = ?
V2 = 2.6 L
T2 = 173 K
Substituting the given values in the equation to get:
P2 = 778.05 mm Hg = 1.02 atm
<span>The steps of solubility of water in N-butanol is as follows:1. N-butanol molecules are attracted to the surface of the water, 2. N-butanol molecules surround water molecules, 3. Butanol mixes with water and 4. Water molecules are carried into N-butanol.</span>
Answer:
Na₂CO₃.2H₂O
Explanation:
For the hydrated compound, let us denote is by Na₂CO₃.xH₂O
The unknown is the value of x which is the amount of water of crystallisation.
Given values:
Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g
Mass after heating (Na₂CO₃) = 3.22g
Mass of the water of crystallisation = (4.31-3.22)g = 1.09g
To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:
Number of moles = 
Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹
Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹
Number of moles of Na₂CO₃ = 
= 0.03mole
Number of moles of H₂O = 
= 0.06mole
From the obtained number of moles:
Na₂CO₃ H₂O
0.03 0.06
Simplest
Ratio 0.03/0.03 0.03/0.06
1 2
Therefore, x = 2
Answer:
The disadvantages of each of the given model of electron configuration have been mentioned below:
1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.
2). Arrow and line diagrams make the counting of electrons and take up too much space.
3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.
Answer:
5.63 mol.
Explanation:
- The balanced chemical equation between NO₂ and H₂O is:
<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g),
</em>
It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.
<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>
<em></em>
- To find the no. of moles of HNO₃ produced:
3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.
8.44 mol of NO₂ produces → ??? mol of HNO₃.
∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.
