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s2008m [1.1K]
4 years ago
10

Material from an environmental study was suspected to contain the banned pesticide, Aldrin, of which you have a pure sample. Gas

chromatography showed 6 peaks. How would you determine unequivocally which of these peaks represented Aldrin?
Chemistry
1 answer:
Anna35 [415]4 years ago
7 0

Answer:

In order  to determine unequivocally which of the peaks represent Aldrin, I would run the pure sample through the chromatography equipment.

Explanation:

Gas chromatography is a technique that separates molecules based on their volatility and interaction with both the stationary phase.

The peaks on the chromatogram show how long a substance took to leave the column. Since each different substance substance will leave the column at a different time, each peak can be attributed to a substance.

Therefore, to know which of the 6 peaks represent Aldrin, it is necessary to run the pure Aldrin in the chromatography equipment and see the time of the peak. Then you just need to compare both chromatograms and indentify Aldrin.

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Help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Rus_ich [418]

Answer:

Tundra

Explanation:

In physical geography, tundra is a type of biome where the tree growth is hindered by low temperatures and short growing seasons.

5 0
3 years ago
Read 2 more answers
50 points!! Brainliest if correct!!
Luda [366]

The answer is C. The answer is C because if u increase the surface area, the more reactants u will get. and if u get more The reactants will move faster. Hoped that Helped!:-)

4 0
3 years ago
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
How does the presence of a catalyst affect the enthalpy of a reaction? Group of answer choices 1. It depends on whether you are
Otrada [13]

Answer:

Option 3. The catalyst does not affect the enthalpy change (\Delta H_\text{rxn}) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction (\Delta H_\text{rxn}) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change \Delta H_\text{rxn} of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction \Delta H_\text{rxn} measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

6 0
3 years ago
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