Answer:
(3) 5.36
Explanation:
Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.
The reaction is:
HAc + NaOH ⇒ NaAc + H₂O
V NaOH = 40 mL x 1 L/1000 mL = 0.040 L
mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol
mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol
mol HAc left after reaction = 0.0025 - 0.002 = 0.0005
Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation
pH = pKa + log ((Ac⁻)/(HAc))
(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)
pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36
THe answer is 5.36
Answer:
Increase
Explanation:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
If the initial temperature and pressure is standard,
Pressure = 1 atm
Temperature = 273.15 K
then we increase the temperature to 400.0 K, The pressure will be,
1 atm / 273.15 K = P₂/400.0K
P₂ = 1 atm × 400.0 K / 273.15 K
P₂ = 400.0 atm. K /273.15 K
P₂ = 1.46 atm
Pressure is also increase from 1 atm to 1.46 atm.
Answer:
Cu(s) in Cu(NO₃)₂(aq)
Explanation:
The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.
The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.
Convert mole to gram by multiplying the molar mass of sodium
0.500mol Na x 22.990g = 11.495g of Na
102 grams.
Equation:
Quantify of heat = mass x specific heat x difference in temperature
We have: quantity of heat : 2300J
specific heat: .449 J/g
difference in t: 80 - 30 = 50
Solve for mass: 2300 = mass x 0.449 x 50
mass = 102.449
2 sig-figs --> 102 grams
