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Brilliant_brown [7]
3 years ago
11

assume that a water balloon has the same number of water molecules as a helium balloon has helium atoms. If the mass of the wate

r is 4.5 times greater than the mass of the helium, how does the mass of a water molecule compare with the mass of a helium atom. please explain
Chemistry
1 answer:
Aloiza [94]3 years ago
3 0

Answer : The mass of the water molecule is 4.5 times greater than the mass of the helium atom.

Explanation :

Assumption : The number of water molecules is equal to the number of helium atoms

Given : The mass of water = 4.5 × The mass of helium     ........(1)

The mass of Water = Mass of 1 water molecule × Number of water molecule

The mass of Helium = Mass of 1 helium atom × Number of helium atom

Now these two masses expression put in the equation (1), we get

Mass of 1 water molecule × Number of water molecule = 4.5 × Mass of 1 helium atom × Number of helium atom

As per assumption, the number of water molecules is equal to the number of helium atoms. The relation between the mass of water molecule and the mass of helium atom is,

Mass of water molecule = 4.5 × Mass of helium atom

Therefore, the mass of the water molecule is 4.5 times greater than the mass of the helium atom.

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Which occurs in the half-reaction Na(s) → Na+
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Answer:

c)

Explanation:

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How to find protons neutrons and electrons of carbon?
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3 years ago
The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5
Sladkaya [172]
<span>a) 7.9x10^9 b) 1.5x10^9 c) 3.9x10^4 To determine what percentage of an isotope remains after a given length of time, you can use the formula p = 2^(-x) where p = percentage remaining x = number of half lives expired. The number of half lives expired is simply x = t/h where x = number of half lives expired t = time spent h = length of half life. So the overall formula becomes p = 2^(-t/h) And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are: a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9 b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9 c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
4 0
3 years ago
Read 2 more answers
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
3 years ago
2. List and elaborate on at least two limitations of the 24-hour recall as a
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Answer:

The correct answer is - may not be typical, and participant burden.

Explanation:

The 24-hour recall is nothing but a retrospective method of diet assessment. In this method, an individual is interviewed about his or her diet consumption during the last 24 hours.

The disadvantages or limitations of this method include the inability of a single day's intake to describe the typical diet, multiple recalls to intake, cost and administration time; participant burden, have to recall to reliably estimate usual intake.

5 0
3 years ago
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