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sertanlavr [38]
3 years ago
13

The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9

04 u, respectively. Compute the average atomic weight of naturally occurring antimony.
Chemistry
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

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Dylan invested $ 600 in a savings account at a 1.6 % annual interest rate . He made no deposits or withdrawals on the account for 2 years. The interest was compounded annually . Find, to the nearest cent, the balance in the account after 2 years. a₁ What information from the question is important ? b . How much does Dylan have in his bank account after 2 years ? Show work .

4 0
2 years ago
Calculate the total pressure in a 10.0 liter flask at 27°C of a sample of gas that contains 6.0 grams of hydrogen, 15.2 grams of
motikmotik

Answer:

The total pressure is 27.8 atm

Explanation:

From the ideal gas equation,

PV = nRT

P (total pressure) = nRT/V

n (total moles of gases) = (6/1 moles of hydrogen) + (15.2/14 moles of nitrogen) + (16.8/4 moles of helium) = 6+1.1+4.2 = 11.3 moles

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P = 11.3×0.082057×300/10 = 27.8 atm

7 0
3 years ago
HELP When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of
babunello [35]

Q=mc*temperature change

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As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is
chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
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