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bekas [8.4K]
2 years ago
5

What are the difficulties encountered in precipitation titration

Chemistry
1 answer:
Mumz [18]2 years ago
8 0

One difficulty encountered in precipitation titration is that it is hard to determine the exact end point of its reaction.

Precipitation titration is a titration in which a  reaction  occurs from the  analyte and titrant  to form an insoluble precipitate.

With the use of silver for the titrations, (argentometric) we are able to develop many precipitation reactions.  

The  precipitation titrimetry  methods with the use of argentometry includes  

• Mohr’s Method

• Fajan’s Method

• Volhard’s Method

Difficulties encountered in precipitation titration includes

  • Getting the exact end point is hard.

  • it is a very slow titration method.

  • it includes periods of filtration and cooling thereby reducing the reactions available for this type of titration.

See more on Precipitation: brainly.com/question/20628792

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Given the following list of densities, which materials would float in a molten vat of lead provided that they do not themselves
vampirchik [111]

Answer:

b. glass and charcoal

Explanation:

Step 1: Given data

  • Density of Pb: 11.4 g/mL
  • Density of Glass: 2.6 g/mL
  • Density of Au: 19.3 g/mL
  • Density of charcoal: 0.57 g/mL
  • Density of platinum: 21.4 g/mL

Step 2: Determine which material will float in molten lead

Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.

8 0
3 years ago
A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne
Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0
3 years ago
Which of the following is not a type of energy
Usimov [2.4K]
C. Heat answer correct that
8 0
3 years ago
Read 2 more answers
The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvin. (Round the final
postnew [5]

Answer:

45K

Explanation:

Rise in temperature = Final - initial temperature.

temperature in K = Temperature in Celsius + 273

for Celsius; T2 -T1 =45°C

for kelvin; T2+273 -(T1+273) = ?

                T2+273 -T1-273 =?

                T2-T1 = ?

               T2-T1 =45k

hence ΔT(K) = ΔT(°C) (temperature difference in Celsius is equal to temperature difference in kelvin)

4 0
3 years ago
T or F: a. A concentrated solution in water will always contain a strong or weak electrolyte
Eduardwww [97]

Answer:

true

Explanation:

8 0
4 years ago
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