Answer is: boiling point will be changed by 4°C.
Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).
Change in boiling point: ΔT =i · Kb · b.
Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.
b - molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.
i - Van't Hoff factor.
Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.
Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.
Answer:
2.12atm
Explanation:
Boyle's Law: P1V1 = P2V2
Manipulate to solve for unknown: P2 = P1V1/V2
Substitute values: P2=(1.2atm)(4.6L)/2.6L
P2 = 2.1230769atm
Round to 3 sig figs to get 2.12atm
Answer:
Last Quarter also called Third Quarter.
Explanation:
The answer is:
the molarity = 50 moles/liters
The explanation:
when the molarity is = the number of moles / volume per liters.
and when the number of moles =2.5 moles
and the volume per liters = 0.05 L
so by substitution:
the molarity = 2.5moles/0.05L
= 50 moles /L
The mass of NaCl formed is 8.307 grams
<u><em> calculation</em></u>
step 1: write the equation for reaction
Na₂CO₃ + 2HCl → 2 NaCl +CO₂ +H₂O
Step 2: find the moles of Na₂CO₃
moles = mass/molar mass
The molar mass of Na₂CO₃ is = (23 x2) + 12 + ( 16 x3) = 106 g/mol
moles = 7.5 g/106 g/mol =0.071 moles
Step 3: use the mole ratio to determine the mole of NaCl
Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl =0.07 x2 =0.142 moles
Step 4: calculate mass of NaCl
mass= moles x molar mass
the molar mass of NaCl= 23 +35.5 =58.5 g/mol
mass = 0.142 moles x 58.5 g/mol =8.307 grams
