Answer:
0.497 moles/L
Explanation:
The reaction that takes place is:
- C₃H₆O₃ + NaOH → C₃H₅ONa + H₂O
First we c<u>alculate the moles of lactic acid in 0,821 g</u>, using its molar mass:
- 0,821 g ÷ 90 g/mol = 9,122x10⁻³mol lactic acid
<em>One mole of lactic acid reacts with one mole of sodium hydroxide</em> (it is a monoprotic acid), so in 18,34 mL of NaOH solution there are 9,122x10⁻³moles of NaOH:
- 18.34 mL ÷ 1000 = 0,01834 L
- 9,122x10⁻³mol ÷ 0,01834 L = 0.497 moles/L
Answer:
0.054 moles
Explanation:
It is the rounded off answer.
Answer:
The heat needed to convert 1 kg of feed water at 20°C into dry saturated steam at a pressure of 9 bar is 2690.19 kJ/kg.
Explanation:
Step 1 : Obtain the enthalpy of staurated steam and enthalpy of evaporation at 9 bar pressure from the steam table:
From steam table, at 9 bar, the enthalpy of saturated water = 742.83 kJ /kg
enthalpy of evaporation = 2031.1 kJ /kg
Step 2: Calculate the enthalpy of dry saturated steam:
Enthalpy of dry saturated steam = enthalpy of saturated water + enthalpy of evaporation
= 742.83 + 2031.1 = 2773. 93 kJ/ kg
Step 3: Calculate the enthalpy of 1 kg of feed water at 20°C
Enthalpy of 1 kg of feed water = c * ( T2 -T1)
= 4.187 * (20-0)
= 83.74 kJ /kg
Step 4 : calculate the heat needed by 1 kg of feed water at 20°C to be converted to dry saturated steam at 9 bar:
Heat needed = Enthalpy of dry saturated steam - enthalpy of feed water
Heat needed = 2773.93 kJ/kg - 83.74 kJ/kg
Heat = 2690.19 kJ/kg
Answer:
18,1 mL of a 0,304M HCl solution.
Explanation:
The neutralization reaction of Ba(OH)₂ with HCl is:
2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O
The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:
= 2,7531x10⁻³moles of Ba(OH)₂
By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:
= 5,5062x10⁻³moles of HCl.
The volume of a 0,304M HCl solution for a complete neutralization is:
= 0,0181L≡18,1mL
I hope it helps!