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Monica [59]
3 years ago
6

Europium has two isotopes. One of the isotopes has a mass of 150.920 amu, and the other has a mass of 152.921 amu. Europium has

an atomic mass of 151.965 amu. What are the percent compositions of the two isotopes
Chemistry
1 answer:
loris [4]3 years ago
5 0

Answer:

The percent compositions of the first isotope = 47.78 %

The percent compositions of the second isotope = 100 - 47.78 % = 52.22 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope :

% = x %

Mass = 150.920 u

For second isotope :

% = 100  - x  

Let, Mass = 152.921 u

Given, Average Mass = 151.965 u

Thus,  

151.965=\frac{x}{100}\times {150.920}+\frac{100-x}{100}\times {152.921}

150.92x+152.921\left(100-x\right)=15196.5

-2.001x=-95.6

Solving for x, we get that:

x=47.78 %

The percent compositions of the first isotope = 47.78 %

The percent compositions of the second isotope = 100 - 47.78 % = 52.22 %

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Answer:

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Explanation:

starch, a white, granular, organic chemical that is produced by all green plants. Starch is a soft, white, tasteless powder that is insoluble in cold water, alcohol, or other solvents.

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3 years ago
What is the correct unabbreviated electron configuration of the following?:
Rudik [331]

The correct  unabbreviated electron configuration  is as below


Vanadium - 1S2 2S2 2P6 3S2 3p6 3d3 4s2

Strontium -  1s2 2S2 2P6 3S2 3P6 3d10 4S2 4P6 4S2

Carbon =1S2 2S2 2P2


<u><em> Explanation</em></u>

vanadium  is in atomic number  23  in the periodic table  hence its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2

Strontium is in atomic number 38  in periodic table  hence  its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d10  4s2 4p6 4s2

Carbon is in atomic number 6 in periodic  table  therefore its electron configuration is 1s2 2s2 2p2

5 0
4 years ago
Did diagram shows embryo development of four different animals. How is this evidence used to suggest that life changes over time
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Octane has a density of 0.692 g/ml at 20oc. how many grams of o2 are required to burn 15.0 gal of c8h18 (the average capacity of
Hatshy [7]

Octane has a density of 0.692 g/ml at 20°C. Grams of O₂ required to burn 15.0 gal of C₈H₁₈ is 1.37868×10⁵g.

Octane is a hydrocarbon having eight carbon atoms and have single bonds only.

Given,

Density of octane = 0.692g/ml

Temperature = 20°C = 293K

Amount of Octane present = 15gal

Molar mass of octane = 114g

We know that,

1 gal = 3.78541L = 3785.4ml

Hence, 15 gal = 56781 ml

Now let's calculate the Mass of octane required:

Mass of octane = 0.692 x 56781

Mass of octane = 39292.45 g

According to the given equation,

C₈H₁₈ + 12.5O₂ ---------> 8CO₂ + 9H₂O

Also, 114g of octane needs-------> 400g of oxygen

39292.45g of octane needs ---------> 137868.24g of oxygen

Hence, oxygen needed to burn octane is 1.37868×10⁵g

Learn more about Octane here, brainly.com/question/21268869

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4 0
2 years ago
A 9.75 L 9.75 L container holds a mixture of two gases at 41 ° C. 41 °C. The partial pressures of gas A and gas B, respectively,
Triss [41]

Answer:

The total  pressure P = 1.642 atm

Explanation:

From the dalton's law

Total pressure of the mixture is

P = P_{A} + P_{B} + P_{C} ----- (1)

P_A = 0.419 atm

P_B = 0.589 atm

P_C = \frac{nRT}{V}

n = 0.24 mole

R = 0.0821 \frac{L.Atm}{K.mol}

T = 41 °c = 314 K

P_C = \frac{(0.24)(0.0821)(314)}{9.75}

P_C = 0.634 atm

From equation (1)

P = 0.419 + 0.589 + 0.634

P = 1.642 atm

Thus the total  pressure P = 1.642 atm

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