The top one is different from the bottom because of is curvature shape while the bottom one is a square shape i think the bottom will heat up faster because of the nice even area inside where heat waves can evenly flow
a. 30 moles of H₂O
b. 2.33 moles of N₂
<h3>Further explanation</h3>
Given
a. 20 moles of NH₃
b. 3.5 moles of O₂
Required
a. moles of H₂O
b. moles of N₂
Solution
Reaction
4NH₃+3O₂⇒2N₂+6H₂O
a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :
=6/4 x mol NH₃
= 6/4 x 20 moles
= 30 moles
b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :
=2/3 x mol O₂
= 2/3 x 3.5 moles
= 2.33 moles
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
1.strong acid: 2
2.weak acid:6
3.strong base:13
4.weak base:8
5.neutral:7
