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3 years ago

6) Which of the following would travel the farthest in column chromatography?

1 answer:

8 0

The answer would be C8H18, or octane. Column chromatography is a separation technique that works off of polarity. The stationary phase, in column chromatography, are non-modified silica gels. Silica gels are normally polar. This means the solvent that will elute first are non-polar molecules...then polar molecules. The polar silica gels have a stronger interaction with polar molecules like CH3OH, and therefore will lag the solvent. Additionally, as you decrease the carbons, in hydrocarbons, the molecule becomes more polar. This is not C6H6 isn't the answer and octane is.

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A liter is the same as a _____.

Nata [24]

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3 years ago

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

1. Table

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

2. Orders

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

3. Rate equation

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

4. Rate constant, k

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

5. Rate law:

$rate=k[A]^2C=0.0075[A]^2[C]$

6. Substitute

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

7 0

3 years ago

What is the energy of a photon emitted with a wavelength of 448 nm?

DIA [1.3K]

Answer:

i think D is correct

Explanation:

so think about it

yourself

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3 years ago

The time necessary for the decay of one- half sample of a radioactive substance is a _____. mass defect daughter half life nucli

Mazyrski [523]

Answer: half life

Explanation: Radioactive decay follows first order kinetics and the time required for the decay of a radioactive material is calculated as follows:

$t=\frac{2.303}{k}\hspace{1mm}log\frac{x}{a}$

t= time required

k= disintegration constant

x= amount of substance left after time t

a= initial amount of substance

when one half of the sample is decayed, one half of the sample remains and t can be represented as $t_{1/2}$

at $t= t_{1/2}$ , $x=\frac{a}{2}$

$t_{1/2}=\frac{2.303}{k}\hspace{1mm}log\frac{a/2}{a}$

$t_{1/2}=\frac{0.693}{k}$

3 0

3 years ago

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The weathered debris in deserts consists mainly of

vampirchik [111]

Answer;

C. unchanged rock and mineral fragments

Explanation;

A large number of landforms and features found in desert environments are formed as the result of weathering. Weathering is defined as the breakdown and deposition of rocks by weather acting in situ

The two main types of weathering which occur in deserts are Mechanical weathering, which is the disintegration of a rock by mechanical forces that do not change the rock's chemical composition and Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.

By contrast much of the weathered debris in deserts has resulted from mechanical weathering. Chemical weathering, however, is not completely absent in deserts. Over long time spans,clays and thin soils do form.

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3 years ago

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