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sineoko [7]
2 years ago
9

If the van urk reagent is used to test a sample and it turns bluish-purple, what drug does this indicate the presence of?

Chemistry
1 answer:
saul85 [17]2 years ago
4 0
 A van urk reagent is considered to be specific and also sensitive of which it is in chromogenic reagents. If it is used in as a test sample, turning the sample in bluish-purple in color, the drug that indicated its own presence is LSD or also known as Lysergic Acid Diethylamide. It is a type of psychedelic drug.
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The mechanism for the reaction described by 2N2O5(g) ---> 4NO2(g) + O2(g) is suggested to be (1) N2O5(g) (k1)--->(K-1) NO2
aliina [53]
Go add me on Snapchat christian3cruz
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2 years ago
Solid calcium carbonate (CaCO3) reacts with hydrochloric acid (HCI) to form carbon dioxide, water, and
Pavlova-9 [17]

Using a more concentrated HCl solution and Crushing the CaCO₃ into a fine powder makes the reaction to occur at a faster rate.

<u>Explanation:</u>

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(aq) + CO₂(g)

When calcium carbonate reacts with hydrochloric acid, it gives out carbon-dioxide in the form of bubbles and there is a formation of calcium chloride in aqueous medium.

The rate of the reaction can be increased by

  • Using a more concentrated HCl solution
  • Crushing the CaCO₃ into a fine powder

When concentrated acid is used instead of dilute acid then the reaction will occur at a faster rate.

When CaCO₃ is crushed into a fine powder then the surface area will increases thereby increasing the rate of the reaction.

3 0
2 years ago
Read 2 more answers
How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 years ago
How would each of the following procedural errors affect the results to be expected in this experiment? Give your reasoning in e
Digiron [165]

Answer:

a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.

b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.

c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.

d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling

temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.

7 0
2 years ago
Rank the following acids in order of increasing acid strength. Key: Weakening of hydrogen bond and stability of resulting anion.
raketka [301]

Explanation:

The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.

This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose H^{+} ion.

Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.

Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.

       HOI < HOBr_{2} < HOCl_{3} < HOF

8 0
2 years ago
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