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3 years ago

A runner competed in a 5-mile run. How many yards did she run?

Chemistry

1 answer:

vfiekz [6]3 years ago

3 0

5 miles are equivalent to 8,800 yards

so b. is the correct answer. plz like and hope it helped

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Choose the term that best fits the following definition: the liquid or gas that carries the sample across the solid support.

iren [92.7K]

The correct answer is option B.

The liquid or gas that carries the sample across the solid support is called mobile phase.

In chromatography, there are two phases: mobile phase and solid phase.

The mobile phase can be either in gas form or liquid form.

While performing the chromatography technique, mobile phase moves over the stationary phase and its components adsorb to the stationary phase and set apart from each other at different rates.

In general, mobile phase refers to the solvent phase that slides over the stationary phase through the chromatography paper.

on the other hand, stationary phase is motionless.

If you need to learn more about gas click here:

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1 year ago

When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Se

Elodia [21]

Answer:

when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.

3 0

3 years ago

The diagram shows the structure of an animal cell.

Vera_Pavlovna [14]

Answer:

Explanation:

The cell membrane is the protective, semi-permeable membrane surrounding a cell and its contents.

8 0

3 years ago

Read 2 more answers

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

How are scientific questions answered?

galben [10]

Answer:

Explanation:

Scientists always perform tests and observe and measure in the physical world to prove their points or answer their questions.

3 0

3 years ago