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Serga [27]
3 years ago
5

A. glycerol

Chemistry
1 answer:
Natasha2012 [34]3 years ago
4 0
Bzjnu8:i9.8&9olzlns soccer
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How many molecules in 3.4 moles of NH4NO3?
aliya0001 [1]

There are  20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.

<h3>How many molecules in 3.4 moles of NH4NO3?</h3>

We know that one mole of a substance has  6.022 × 10²³ molecules so in 3.4 moles of NH4NO3, we have 20.5 x 10^24 molecules if we multiply the  6.022 × 10²³ with 3.4.

So we can conclude that there are  20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.

Learn more about mole here: brainly.com/question/15356425

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5 0
1 year ago
Observation And Assessment
Sloan [31]
I'm confused here??????
5 0
3 years ago
Which molecule plays the greatest role in the thermal regulation of the troposphere?
makkiz [27]

Answer: The moon was 20 farms away from earth so If I ate a couch then went to my treadmile outside of my cow, I could add the letter Grammar to my checklist.

Explanation: You see, If there was a man who had no arms or legs he would be walking with his feet right? SO If I added chickens to my head and ate the eifel tower am I 50% a choclate bar. Hope this helps

5 0
2 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
3 years ago
. If the half-life of cesium-133 is 30 years, how much of a 600g sample of cesium-133 will be left after
e-lub [12.9K]

Answer:

Caesium (55Cs) has 40 known isotopes, making it, along with barium and mercury, one of the elements with the most isotopes. The atomic masses of these isotopes range from 112 to 151. Only one isotope, 133Cs, is stable. The longest-lived radioisotopes are 135Cs with a half-life of 2.3 million years, ... It constitutes most of the radioactivity still left from the Chernobyl accident ...

3 0
2 years ago
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