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Lesechka [4]
4 years ago
8

In Atomic Absorption Spectrophotometry, narrow atomic lines are desirable. However, line broadening arises from: (a) Jablonski e

ffect (b) Doppler effect (c) Plasma Effect (d) Background effect (e) Blue sky effect
Chemistry
1 answer:
LenKa [72]4 years ago
7 0

Answer: b) Doppler Effect

Explanation:

Line broadening in AAS, arises due to some effects, which can occur due to a number of factors. The line width broadening effects include: Doppler, Lorentz, Self absorption, and quenching effects.

Doppler effect arises because along the line of observation, atoms will have different components of velocity.

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A student practicing for a track meet ran 300 meters in 30 seconds. what was her average speed?​
ololo11 [35]

Answer:

<h2>10 m/s</h2>

Explanation:

Her average speed can be found by using the formula

v =  \frac{d}{t}  \\

d is the distance

t is the time taken

From the question we have

v =  \frac{300}{30}  = 10 \\

We have the final answer as

<h3>10 m/s</h3>

Hope this helps you

7 0
3 years ago
Which element is in group 2 and period 7 of the periodic table
bekas [8.4K]
Hey
      Sorry, But There Is No 7th Element in group two. only six which is Radium.
     ~Spades15
3 0
4 years ago
When he carried out his experiment, Nathan found that he had three small
Ann [662]

I think it's either B or D

Explanation:

it really depends on how differently shaped the beakers are sorry if this doesn't help very much

3 0
3 years ago
Yoooo help me w this one
Scilla [17]

Answer:

0.84kg of gatorade powder

Explanation:

From the question given, we were told that 0.6kg of gatorade powder required 5 gallons of water.

To obtain the mass of gatorade needed for 7 gallons of water, we simply do the following:

0.6kg of gatorade powder required 5 gallons of water.

Therefore, xkg of gatorade powder will require 7 gallons of water i.e

xkg of gatorade powder = (0.6 x 7)/5

xkg of gatorade powder = 0.84kg

Therefore, 0.84kg of gatorade powder will be required.

6 0
3 years ago
If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Cels
Gre4nikov [31]

Answer:

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹  → 400K

We convert the T° to C → 400 K - 273 = 127°C

4 0
3 years ago
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