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andriy [413]
2 years ago
14

Determine the volume in mL of 0.37 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 2

0.8 mL of 0.51 M CH3CH2NH2(aq). Enter your answer with one decimal place. The Kb of ethylamine is 6.5 x 10-4.
Chemistry
1 answer:
Dimas [21]2 years ago
7 0

Answer:

14.3mL you require to reach the half-equivalence point

Explanation:

A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:

CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.

Initial moles of CH₃CH₂NH₂ are:

20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =

0.0106moles CH₃CH₂NH₂

To reach the half-equivalence point you require:

0.0106moles ÷ 2 = 0.005304 moles HClO₄

As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:

0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =

<h3> 14.3mL you require to reach the half-equivalence point</h3>

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Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g
Andreas93 [3]
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
7 0
3 years ago
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user100 [1]
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7 0
2 years ago
If the natural abundance of Ag-107 is 51.84% what is the abundance of Ag-109?
ASHA 777 [7]

Answer:

Explanation:

48.16%

Well, both abundances have to total 100% so is Ag-107 is 51.84%, then Ag-109 must be 100 – 51.84 = 48.16%.

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4 0
2 years ago
A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C
laila [671]

Answer:

  • First choice: 99 J/K

Explanation:

<u>1) First law of thermodynamic (energy balance)</u>

  • Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

<u>2) Energy change of each substance:</u>

  • General formula:

     

Heat released or absorbed = mass × Specific heat × change in temperature

  • density of water: you may take 0.997 g/ ml as an average density for the water.

  • mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

  • Specif heat of water: 1 cal / g°C

  • Heat released by the hot water:

       Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

  • Heat absorbed by the cold water:

       Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

  • Heat absorbed by the calorimeter

       Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

<u>4) Balance</u>

  • Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

  • Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

  • Convert to cal / K to Joule / K

  • 1 cal = 4.18 Joule

       23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0
3 years ago
2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)
defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

                        2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

                       207.2 g of Pb ----------------- 1 mol

                             x g             ----------------- 0.0013 moles

                        x = (0.0013 x 207.2) / 1

                        x = 0.269 g of Pb                                                                

8 0
3 years ago
Read 2 more answers
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