The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
Answer:
= 913.84 mL
Explanation:
Using the combined gas laws
P1V1/T1 = P2V2/T2
At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.
V1 = 80.0 mL
P1 = 109 kPa
T1 = -12.5 + 273 = 260.5 K
P2 = 10 kPa
V2 = ?
T2 = 273 K
Therefore;
V2 = P1V1T2/P2T1
= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)
<u>= 913.84 mL</u>
Answer:
0.077 M
Explanation:
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
<span>A beryllium atom has 4 electrons.
1, 0, 0, +1/2
1, 0, 0, -1/2
2, 0, 0, +1/2
2, 0, 0, -1/2</span>