Answer:
u should call me justin bc I was just in your mom
Answer:
1.) 0.1 M
2.) 0.2 M
3.) 1 M
4.) Solution #3 is the most concentrated because it has the highest molarity. This solution has the largest solute to solvent ratio. The more solvent there is, the lower the concentration and molarity.
Explanation:
To find the molarity, you need to (1) convert grams NaOH to moles (via molar mass from periodic table) and then (2) calculate the molarity (via the molarity equation). All of the answers should have 1 sig fig to match the given values.
Molar Mass (NaOH): 22.99 g/mol + 16.00 g/mol + 1.008 g/mol
Molar Mass (NaOH): 39.998 g/mol
4 grams NaOH 1 mole
---------------------- x ------------------ = 0.1 moles NaOH
39.998 g
1.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (1 L)
Molarity = 0.1 M
2.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.5 L)
Molarity = 0.2 M
3.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.1 L)
Molarity = 1 M
Answer:
2.6 ×10^-42
Explanation:
From
∆G= -RTlnK
∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1
R= 8.314 Jmol-1K-1
T= 25°C + 273= 298K
-237.2×10^3= 8.314 × 298 × ln K
ln K= -237.2×10^3/2477.572
K = 2.6 ×10^-42
The pH of the solution after adding 0.150 moles of solid LiF is 3.84
<u>Explanation:</u>
We have the chemical equation,
HF (aq)+NaOH(aq)->NaF(aq)+H2O
To find how many moles have been used in this
c= n/V=> n= c.V
nHF=0.250 M⋅1.5 L=0.375 moles HF
Simillarly
nF=0.250 M⋅1.5 L=0.375 moles F
nHF=0.375 moles - 0.250 moles=0.125 moles
nF=0.375 moles+0.250 moles=0.625 moles
[HF]=0.125 moles/1.5 L=0.0834 M
[F−]=0.625 moles/1.5 L=0.4167 M
To determine the problem using the Henderson - Hasselbalch equation
pH=pKa+log ([conjugate base/[weak acid])
Find the value of Ka
pKa=−log(Ka)
pH=−log(Ka) +log([F−]/[HF]
pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)
pH=-log(3.5 x 10 ^4)+log(4.996)
pH= -4.54+0.698
pH=-(-3.84)
pH=3.84
The pH of the solution after adding 0.150 moles of solid LiF is 3.84
Answer:
RbF
mgo
nh4cl
because electrons are lost by and element forming a cation and gained by the other element forming an anion and held together by electrostatic forces
