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2 years ago

What is A collection of waves called

1 answer:

5 0

When waves act together, you talk about "interference". When they reinforce each other, it is "constructive interference". When they cancel each other, it is "destructive interference".

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O efeito de um inibidor é...

max2010maxim [7]

Fśa méj hšvâśhi porquse

6 0

2 years ago

The Lyman series results from excited state hydrogen atoms transiting to

Nutka1998 [239]

Answer:

I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.

n = 5 4th excited state

n = 4 3rd excited state

n = 3 2nd excited state

n = 2 1st excited state

n = 1 ground state

Here are the possible spectral lines.

n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.

n = 4 to 3, 4 to 2, 4 to 1 = 3 lines

n = 3 to 2, 3 to 1 = 2 lines

n = 2 to 1 = 1 line. Add 'em up. I get 10.

b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.

c.The E for any level is -21.8E-19 Joules/n^2

To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.

So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.

Explanation:

8 0

2 years ago

It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at const

lilavasa [31]

Given :

Energy , E = 330 J .

Initial temperature , $T_i=21^oC$ .

Final temperature , $T_f=24.6^oC$ .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

$n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol$

Energy required is given by :

$q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}$

Hence , this is the required solution .

8 0

2 years ago

A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1

icang [17]

Answer:

Approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

$\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C$ .

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took $2911\; \rm J$ of energy to raise the temperature of this sample by $\Delta T = 92.6\; \rm ^\circ\! C$ . Therefore, raising the temperature of this sample by $1\; \rm ^\circ\! C$ (unit temperature) would take only $\displaystyle \frac{1}{92.6}$ as much energy. That corresponds to approximately $31.436\; \rm J$ of energy.

On the other hand, the energy required to raise the temperature of this material by $1\; \rm ^\circ\! C$ is proportional to the mass of the sample (also assuming no phase change.)

It took approximately $31.436\; \rm J$ of energy to raise the temperature of $57.07\; \rm g$ of this material by $1\; \rm ^\circ C$ . Therefore, it would take only $\displaystyle \frac{1}{57.07}$ as much energy to raise the temperature of $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C\!$ . That corresponds to approximately $0.551\; \rm J$ of energy.

In other words, it takes approximately $0.551\; \rm J$ to raise $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C$ . Therefore, by definition, the specific heat of this material would be approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

8 0

2 years ago

Write the name for each of the following chemical formulas. <br> d. P3N5 (for doping semiconductors)

rewona [7]

Answer:

Triphorphorus Pentanitride

Explanation:

since there are three phosphorus use the prefix tri-, since there are five nitrogen use the prefix penta. tri=3, penta=5. then add a suffix -ide to the end of nitrogen. that will give you triphosphorus pentanitride.

3 0

1 year ago