Answers
Answer:
0.1775 M
Explanation:
The reaction that takes place is:
- HA + NaOH → NaA + H₂O
Where HA is the unknown weak acid.
At the equivalence point all HA moles are converted by NaOH. First we calculate how many NaOH moles reacted, using the given concentration and volume:
- 0.0500 M NaOH * 14.2 mL = 0.71 mmol NaOH
That means that in 4.00 mL of the weak acid solution, there were 0.71 weak acid mmoles. With that in mind we can now calculate the concentration:
- 0.71 mmol HA / 4.00 mL = 0.1775 M
Related Questions
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2Cl-(aq)--->Cl2(g)+2e-
Cr3+(aq)+3e- ---->Cr(s)
What is the final, balanced equation for this reaction?
1.) 2cr3+(aq)+6Cl-(aq) ------> 2Cr(s)+3Cl2(g)
2.) 2Cr3(aq)+2Cl-(aq)+6e- --->Cl2(g)+2Cr(s)
3.) Cr3+(aq)+6Cl-(aq)+3e- ---->2Cr(g)+3Cl2(g)
4.) Cr3+(aq)+2Cl-(aq)------>Cr(s)+Cl2(g)
Answers
Explanation:
1) Write the oxidation half-reaction:
2) Write the reduction half-raction:
3) Multiply each half-reaction by the appropiate coefficient to equal the number of electrons of both half-reactions.
4) Add both half-reactions
And that is the answer. You can count the atoms and charges on every side and check they are equal.
Answer:
A
Explanation:
got it correct on edge
Answers
Answer
Naphthalene is a non electrolyte
If the unknown compound is an electrolyte it gives 2 or more ions in solution
( NaCl >> Na+ + Cl- => 2 ions
Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)
the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )
For naphthalene
delta T = 1.86 x m
for a salt that gives 2 ions
delta T = 1.86 x m x 2
hence the lowering in freezion point of unkown is greater then napthalene
(B) H2S
(C) SO3
(D) PCl3
(E) CH2Cl2
(F) NO2
Answers
Answer:
D and E
Explanation:
D. PCl3
E. CH2Cl2
Answers
Answer:
35.6 g of W, is the theoretical yield
Explanation:
This is the reaction
WO₃ + 3H₂ → 3H₂O + W
Let's determine the limiting reactant:
Mass / molar mass = moles
45 g / 231.84 g/mol = 0.194 moles
1.50 g / 2 g/mol = 0.75 moles
Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.
Let's make rules of three:
1 mol of tungsten(VI) oxide needs 3 moles of H₂
Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)
3 moles of H₂ need 1 mol of WO₃ to react
0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles
It's ok. I do not have enough WO₃.
Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.
Let's convert the moles to mass (molar mass . mol)
0.194 mol . 183.84 g/mol = 35.6 g
Answers
Answer: Yes the absorb in the visible range.
Explanation:
The relationship between wavelength and energy of the wave follows the equation:
where,
= energy of the wave = 182 kJ/mol = 182000 J/mol
N = avogadro's number =
h = plank constant =
c = speed of light =
= wavelength of the wave = ?
Putting all the values:
The wavelength range for visible rays is 400 nm to 750 nm, thus the complex absorb in the visible range.
Final answer:
The expected wavelength of the absorption in the [CrF6]3- complex cannot be calculated without knowing the exact value of ΔE. Therefore, it is unclear if the complex will absorb in the visible range.
Explanation:
In this complex, the value of delta (Δ) is 182 kJ/mol. To calculate the expected wavelength of the absorption corresponding to the promotion of an electron from the lower-energy orbital to the higher-energy orbital, we can use the equation: Wavelength = (hc)/(ΔE), where h is Planck's constant and c is the speed of light. By substituting the given values and dividing by Avogadro's number, we can calculate the expected wavelength.
However, the information given in the question does not provide the exact value for ΔE. Without this information, it is not possible to calculate the expected wavelength accurately. Therefore, we cannot determine if the complex will absorb in the visible range.
Learn more about Calculating expected wavelength of absorption in a complex here:
#SPJ3
Answers
Answer:
ΔHrxn = 193107.69 J/mol
Explanation:
ΔHrxn = mcΔT
m = mass
c = heat capacity
ΔT = temperature variation
density = m/V
m = density x V
m = 1.00 g/mL x 400.0 mL
m = 400.0 g
ΔHrxn = mcΔT
ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C
ΔHrxn = 10041.6 J
CaO + 2HCl → CaCl₂ + H₂O
CaO = 56.0774 g/mol
2.90 g CaO = 0.052 mol
400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl
ΔHrxn = 10041.6 J is for 0.052 mol of CaO
ΔHrxn = 193107.69 J is for 1 mol of CaO