One of the best buffer choice for pH = 8.0 is Tris with Ka value of 6.3 x 10^-9.
To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value:
pKa = -log Ka
For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
pKa = -log Ka
= -log (6.3x10^-9)
= 8.2
We know that buffers work best when pH is equal to pKa:
pKa = 8.2 = pH
Therefore Tris would be a best buffer at pH = 8.0.
17.8 mL NaOH
<em>Step 1.</em> Write the chemical equation
Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)
<em>Step 2.</em> Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]
= 11.50 mmol Fe^(2+)
<em>Step 3.</em> Calculate the moles of NaOH
Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]
= 23.00 mmol NaOH
<em>Step 4.</em> Calculate the volume of NaOH
Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)
= 17.8 mL NaOH
Answer:
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure 2.2.2 ). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Explanation:
Answer:
2) 0.4 mol
Explanation:
Step 1: Given data
- Volume of the solution (V): 500 mL
- Molar concentration of the solution (M): 0.8 M = 0.8 mol/L
Step 2: Convert "V" to L
We will use the conversion factor 1 L = 1000 mL.
500 mL × 1 L/1000 mL = 0.500 L
Step 3: Calculate the moles of KBr (solute)
The molarity is the quotient between the moles of solute (n) and the liters of solution.
M = n/V
n = M × V
n = 0.8 mol/L × 0.500 L = 0.4 mol
Answer: 7.07 grams
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 1 mole of 
Thus 0.052 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 1 mole of 
Thus 0.052 moles of give = of
Mass of 
Thus 7.07 g of will be produced from the given masses of both reactants.
