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astraxan [27]
3 years ago
6

A student was performing a separation of a mixture of organic compounds. The final step of the process involved a filtration of

the analyte from an aqueous solution. After drying the filtered solid for a very short period time, they took the melting point of the compound. The measured melting point range of the compound was 106 – 113.8 0C, while the literature melting point of the compound was 122.3 0C. The above scenario is a very common one in organic labs.
1. Do you think their sample was pure?
2. If not, then what do you think could be the source of error.
3. How do you think this error can be minimized?
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer:

1) No

2) The solvent contaminated the analyte

3) The solvent should be evaporated properly before washing and drying the analyte

Explanation:

During separation of organic compounds, solvents are used. These solvents are able to contaminate the analyte and lead to a large difference in melting point of solids obtained.

However, the error can be minimized by evaporating the solvent before washing, drying and melting point determination of the solid.

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Brainliest if correct!!!!!
BlackZzzverrR [31]

Answer:

A

Explanation:

4 0
2 years ago
Read 2 more answers
The balanced equation for the reaction of ammonia and oxygen is the following. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The stand
ale4655 [162]

Answer:

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

7 0
3 years ago
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s
Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 0.0632

t = time taken for the process  = ?

[A_o] = initial amount or concentration of the reactant  = 1.28 M

[A] = amount or concentration left time 't' = 1.28\times \frac{17}{100}=0.2176M

Now put all the given values in above equation, we get:

0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}

t=28s

Therefore, the time taken for the reaction is, 28 s.

8 0
3 years ago
A sample of TNT, C7H5N3O6 , has 8.94 × 1021 nitrogen atoms. How many hydrogen atoms are there in this sample of TNT?
Bess [88]

Answer:

1.49×10²² atoms of H are contained in the sample

Explanation:

TNT → C₇H₅N₃O₆

1 mol of this has 7 moles of C, 5 moles of H, 3 moles of N and 6 moles of O

Let's determine the mass of TNT.

Molar mass is = 227 g/mol

As 1 mol has (6.02×10²³) NA atoms, how many moles are 8.94×10²¹ atoms.

8.94×10²¹ atoms / NA = 0.0148 moles

So this would be the rule of three to determine the mass of TNT

3 moles of N are in 227 g of compound

0.0148 moles of N are contained in (0.0148 .227) / 3 = 1.12 g

Now we can work with the hydrogen.

227 grams of TNT contain 5 moles of H

1.12 grams of TNT would contain (1.12 .5) / 227 = 0.0247 moles

Finally let's convert this moles to atoms:

0.0247 mol . 6.02×10²³ atoms / 1 mol = 1.49×10²² atoms

8 0
3 years ago
A container holds 6.4 moles of gas. Hydrogen gas makes up 25% of the total moles in the container. If the total pressure is 1.24
Degger [83]
  The   partial  pressure of hydrogen is 0.31  atm

calculation

find the number of  hydrogen   moles the container, that is

25/100  x 6.4  =1.6 moles of hydrogen

find the  partial pressure for hydrogen  in 1.6 moles

that is   6.4  moles=  1.24 atm
            1.6  moles= ?

by  cross  multiplication

1.6moles  x1.24  atm/ 6.4 moles=  0.31 atm
6 0
3 years ago
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