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3 years ago

Which of the following is altered by a catalyst?

1 answer:

4 0

In chemistry, a catalyst can speed up the reaction (or make it initiate easier) by altering the activation energy, lowering it enough to allow the reactants to react more easily. Some negative catalysts or inhibitors can do the same by increasing the activation energy.

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

3 years ago

Distinguish between dunes and loess

agasfer [191]

Answer:

Dunes are loose sand hills.

Loess is a compacted wind-blown formation of sediment.

Explanation:

Sand dunes are, as their name implies, made of sand which is itself made of tiny quartz pieces. Sand dunes are present where there is a ready source of broken down sandstone or other type of rock and wind to transport the sand. The dunes are mobile and loose and migrate over time.

Loess comes in vast formations and was created in a very different way. In the last ice age, winds from the north carried sediments loosened and ground by glaciers southwards. There, the sediments accumulated in large formations and became compacted over time.

6 0

3 years ago

What is missing from this graph? A. axis labels B. a trendline C. a title

Iteru [2.4K]

Answer:

Axis Labels

Explanation:

The axis labels are usually located on the x and y axis. This graph however is missing those.

hope this helps!

7 0

3 years ago

How does changing the concentration of the reactants change the parts of a rate law? It changes the rate, R. It changes the rate

Sidana [21]

Answer

A. It changes the rate, R

Explanation

When we change the concentration of the reactants in a chemical reaction, it affects the rate of reaction that happens in the process. Typically, the rate of reaction will decrease with time if the concentration of the reactants decreases because the reactants will be converted to products. Similarly, the rate of reaction will increase when the concentration of reactants are increased.

4 0

3 years ago

If the natural abundance of Ag-107 is 51.84% what is the abundance of Ag-109?

ASHA 777 [7]

Answer:

Explanation:

48.16%

Well, both abundances have to total 100% so is Ag-107 is 51.84%, then Ag-109 must be 100 – 51.84 = 48.16%.

Hope This Helps :)

4 0

3 years ago