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Sedbober [7]
3 years ago
6

The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances

of 39K and 41k are 93.2581% and 6.7302%, respectively. Determine the isotopic mass of 41K.
Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
8 0

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:  

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu  

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 +  X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

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4.A 100 L sample of gas is at a pressure of 80 kPa and a temperature of 200 K. What volume does the same
dexar [7]

Answer:

V₂ = 107.84 L

Explanation:

Given data:

Initial volume = 100 L

Initial pressure = 80 KPa (80/101 =0.79 atm)

Initial temperature = 200 K

Final temperature =273 K

Final volume = ?

Final pressure = 1 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁T₂  /T₁P₂

V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm

V₂ =21567 atm.L.K /200 K.atm

V₂ = 107.84 L

8 0
3 years ago
How much potassium chlorate would dissolve in 200g water at 20°C?​
DedPeter [7]

Answer:

14 solubility in 200 gram of water at 20 c

Explanation:

Hope this helped, and pleas mark as Brainliest :)

6 0
2 years ago
Hess’s Law can be used to determine the heat of reaction in the lab. True False
Feliz [49]
I think it's false I'm just guessing
4 0
3 years ago
Calculate the equilibrium constant for the decomposition of water 2h2o(l)  2h2(g) + o2(g) at 25°c, given that g°f (h2o(l)) = –
kow [346]

Answer:

2.6 ×10^-42

Explanation:

From

∆G= -RTlnK

∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1

R= 8.314 Jmol-1K-1

T= 25°C + 273= 298K

-237.2×10^3= 8.314 × 298 × ln K

ln K= -237.2×10^3/2477.572

K = 2.6 ×10^-42

3 0
3 years ago
How many double bonds does CCL2H2 have?
galben [10]
None. Both chlorines and both hydrogens are single-bonded to the central carbon atom; the molecule is comprised of four single bonds and no double bonds.

Hope this helps!
4 0
2 years ago
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