Answer:
CuSO4 + 2 NaOH = Cu(OH)2 + Na2SO4
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
A
Explanation:
What the equation is tell you is that for every 3 mols of NO2 you get 2 mol of HNO3
3 mol NO2 / 2 mol HNO2 ===> 300.00 mol NO2 / x Cross multiply
3x = 2 * 300
3x = 600 Divide by 3
3x/3 = 600/3 Do the division
x = 200.00
Answer:
Explanation:
L
=
1.10
L
of solution
Explanation:
The Molarity
M
is calculated by the equation comparing moles of solute to liters of solution
M
=
m
o
l
L
For this question we are given the Molarity 0.88M
We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride
We can convert the mass of LiF to moles by dividing by the molar mass of LiF
Li = 6.94
F = 19.0
LiF = 25.94 g/mole
25.2
g
r
a
m
s
x
1
m
o
l
25.94
g
r
a
m
s
=
0.97
moles
Now we can take the the molarity and the moles and calculate the Liters of solution
M
=
m
o
l
L
M
L
=
m
o
l
L
=
m
o
l
M
L
=
0.97
m
o
l
0.88
M
L
=
1.10
L
of solution i just did look at my papaer
Squeezing just the juices out of the orange, like with your hand or whatever you use, is a physical change. yes :)
