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Elza [17]
3 years ago
12

Boron has an average mass of 10.81. One isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent. The

other isotope has a relative abundance of 80.20 percent. What is the mass of the two isotopes?
Chemistry
1 answer:
Andrej [43]3 years ago
5 0

The average mass of an atom is calculated with the formula:

average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ...  an so on

For the boron we have two isotopes, so the formula will become:

average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)

We plug in the values:

10.81 = 0.1980 × 10.012938  + 0.8020 × mass of isotope (2)

10.81 = 1.98 + 0.8020 × mass of isotope (2)

10.81 - 1.98 = 0.8020 × mass of isotope (2)

8.83 = 0.8020 × mass of isotope (2)

mass of isotope (2) = 8.83 / 0.8020

mass of isotope (2) = 11.009975

mass of isotope (1) = 10.012938 (given by the question)

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Answer:

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True

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1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

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2.-

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Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

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Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

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