Answer:
2.40 M
Explanation:
The molarity of a solution tells you how many moles of solute you get per liter of solution.
Notice that the problem provides you with the volume of the solution expressed in milliliters,
mL
. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor
1 L
=
10
3
mL
Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,
g mol
−
1
, which means that you're going to have to convert the mass of the sample from milligrams to grams
1 g
=
10
3
mg
Sodium chloride,
NaCl
, has a molar mass of
58.44 g mol
−
1
, which means that your sample will contain
unit conversion
280.0
mg
⋅
1
g
10
3
mg
⋅
molar mass
1 mole NaCl
58.44
g
=
0.004791 moles NaCl
This means that the molarity of the solution will be
c
=
n
solute
V
solution
c
=
0.004791 moles
2.00
⋅
10
−
3
L
=
2.40 M
The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.
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.
Answer:
Explanation:
Hello there!
In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:
It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:
Best regards!
Answer:
a) pH = 4.68 (more effective)
b) pH =4.44.
Explanation:
The pH of buffer solution is obtained by Henderson Hassalbalch's equation.
The equation is:
![pH =pKa +log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3DpKa%20%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
a) pKa of acetic acid = 4.74
[salt] = [CH₃COONa] = 1.4 M
[acid] = [CH₃COOH] = 1.6 M
This is more effective as there is very less difference in the concentration of salt and acid.
b) pKa of acetic acid = 4.74
[salt] = [CH₃COONa] = 0.1 M
[acid] = [CH₃COOH] = 0.2 M
