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sdas [7]
3 years ago
15

Wavelength (nm)n2E (J) 404.7 435.8 546.1 579.0 Show your work (or send in a separate sheet with your work on it): Q1: A sodium v

apor lamp is similar to a mercury vapor lamp. Sodium displays a single visible emission line at 589.3 nm. Why is it better to use a mercury vapor lamp for calibration purposes
Chemistry
1 answer:
lora16 [44]3 years ago
7 0

Answer: Please see answer below

Explanation:

Mecury vapor lamp is better to use than Sodium vapor light, this is because  because

---The Filaments of the lamp in sodium  emit fast moving electrons, which causes valence electrons of the sodium atoms to excite to higher energy levels which when electrons after being excited, relax by emitting yellow light which concentrates on the the  monochromatic bright yellow part of the visible spectrum which is about  580-590 or about (589nm) which will fall incident on the calibrations making it difficult to see

While

In Mercury vapor lamp, The emitted  electrons from the filaments, after having been excited  by high voltage, hit the mercury atoms but the excited electrons of mercury atoms relax and emits an  ultraviolet  uv invisible lights falling on the mecury vapour lamp to produce white light  covering  a wide range of  (380-780 nm) which is visible that is why it is used for calibrations purposes in  lightening applications.

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Answer:

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3 years ago
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6 0
3 years ago
A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24
SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0
3 years ago
How to solve what volume of 12.0 m hcl is required to make 75.0 ml of 3.50 m hcl?
pochemuha
75mL=0.075L

Molarity=moles/Liters
Moles=0.075(3.50)
=0.2625

Since the number of moles never changes,
we know there are 0.2625moles (or 0.26) in the 12M HCl.
0.2625/12=0.021875L

(0.022L)
4 0
3 years ago
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