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myrzilka [38]
3 years ago
14

Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction.

Chemistry
1 answer:
krok68 [10]3 years ago
5 0
You didn't include much information as to what you were referring too, however I answered a similar question recently to a friend so I'm hoping this is the same question in complete as to yours, if not I apologize.

a.) 2Na(s) + Cl2(g) -----> 2NaCl(s) 
Na --> Na+1 is oxidized 
Cl2 --> Cl- is reduced 

b.) 2HNO3(aq) + 6HI(aq) ----> 2NO(g) + 3I2(s) + 4H2O(l) 
N+5 --> N+2 is reduced 
I-1 --> I2 is oxidized 
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What was the effect of decreasing the copper's mass?
kondor19780726 [428]

Answer:

Increasing the temperature of the copper made the final temperature increase and decreasing the temperature of the copper made the final temperature decrease. ... How does changing the initial mass of the copper affect how much heat energy it has? The more copper, the more heat energy.

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If the detector is capturing 3.3×108 photons per second at this wavelength, what is the total energy of the photons detected in
Reil [10]

Answer:

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

Explanation:

The energy carried by electromagnetic radiation is displaced by waves. This energy is not continuous, but is transmitted grouped into small "quanta" of energy called photons. The energy (E) carried by electromagnetic radiation can be measured in Joules (J). Frequency (ν or f) is the number of times a wave oscillates in one second and is measured in cycles / second or hertz (Hz). The frequency is directly proportional to the energy carried by a radiation, according to the equation: E = h.f, (where h is the Planck constant = 6.63 · 10⁻³⁴ J / s).

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. it is expressed in units of length (m). In light and other electromagnetic waves that propagate at the speed of light (c), the frequency would be equal to the speed of light (≈ 3 × 10⁸ m / s) between the wavelength :

f=\frac{speed of light}{wavelength}

So:

E=\frac{h*speed of light}{wavelength}

In this case, the wavelength is 3.35mm=3.35*10⁻³m and the energy per photon is:

E=\frac{6.63*10^{-34}*3*10^{8}}{3.35*10^{-3} }

E=5.93*10⁻²³ \frac{J}{proton}

The detector is capturing  3.3*10⁸ photons per second. So, in 1 hour:

E=5.93*10^{-23} \frac{J}{proton} *3.3*10^{8} \frac{proton}{s} *\frac{60}{1} \frac{s}{minute} *\frac{60}{1} \frac{minute}{hr}

E=7.04*10⁻¹¹ \frac{J}{hr}

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

3 0
3 years ago
Can anyone help me on this please?
Bad White [126]
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one.  You can't
bluff through it.


7 0
3 years ago
HELPPP
Ksivusya [100]
3 I’m pretty sure maybe
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