The molarity of acetic acid in the vinegar is 0.94 M
<u><em> calculation</em></u>
Step 1: write the balanced equation between CH3COOH + NaOH
that is CH3COOH + NaOH → CH3COONa + H2O
step 2 : find the moles of NaOH
moles =molarity x volume in L
volume in liters = 26.90/1000=0.0269 l
moles = 0.175 mol /L x 0.0269 L =0.0047 moles of NaOH
Step 3: use the mole ratio to find moles of CH3COOH
that is the mole ratio of CH3COOH: NaOH is 1:1 therefore the moles of CH3COOH is =0.0047 moles
Step 4: find the molarity of CH3COOH
molarity = moles/volume in liters
volume in liter = 5.00/1000 =0.005 l
molarity is therefore=0.0047 moles/ 0.005 l = 0.94 M
The energy is transformed into kinetic energy which makes the substance to move. The law of conservation of energy which is the first law of thermodynamics states that in a closed system energy can neither be created nor destroyed but can change from one form to another
Answer:
9.9 ml of 0.200M NH₄OH(aq)
Explanation:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
?ml of 0.200M NH₄OH(aq) reacts completely with 12ml of 0.550M FeCl₃(aq)
1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution
1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)
=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters
The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
