The answer is 4.9 moles.
Solution:
Using the equation for boiling point elevation Δt,
Δt = i Kb m
we can rearrange the expression to solve for the molality m of the solution:
m = Δt / i Kb
Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol,
m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
= 4.883 mol/kg
From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:
m = number of moles / 1.0kg
number of moles = m*1.0kg
= (4.883 mol/kg) * (1.0kg)
= 4.9 moles
Saliva's buffering capacity and flow of secretion are directly related to the rate and extent of demineralization. ... Saliva can act as a replenishing source and inhibit tooth demineralization during periods of low pH, while promoting tooth remineralization when the pH returns to a neutral state.
Answer:
16mL
Explanation:
Using the following formula;
CaVa = CbVb
Where;
Where
Ca = concentration/molarity of acid (M)
Va = volume of acid (mL)
Cb = concentration/molarity of base (M)
Vb = volume of base (mL)
According to the information provided in this question;
Ca (HCl) = 2M
Cb (NaOH) = 5M
Va (HCl) = 40mL
Vb (NaOH) = ?
Using CaVa = CbVb
Vb = CaVa/Cb
Vb = 2 × 40/5
Vb = 80/5
Vb = 16mL
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
The correct answer is option 3. The IUPAC name is Iron(II) sulfide. It is the less stable amorphous form. When this is powdered, it is pyrophoric or it ignites spontaneously in air. It readily reacts with hydrochloric acid producing hydrogen sulfide.
