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In-s [12.5K]
4 years ago
8

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium (Sg) in honor of Glenn T. Seaborg, discov

erer of the transuranium elements. ²⁶³Sg was produced by the bombardment of ²⁴⁹Cf with a beam of ¹⁸O nuclei. Complete and balance a nuclear equation for this reaction.

Chemistry
1 answer:
jolli1 [7]4 years ago
3 0

Answer:

The balanced equations are in the attached file below.

Explanation:

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3 years ago
A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion as b
AnnyKZ [126]

Answer : The mass percentage of barium in the compound is, 53.8 %

Explanation : Given,

Mass of barium compound = 441 mg

Mass of barium sulfate = 403 mg = 0.403 g       (1 mg = 0.001 g)

The balanced chemical reaction will be:

Ba^{2+}(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2Na^+(aq)

First we have to calculate the moles of BaSO_4

\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}

Molar mass of BaSO_4 = 233.38 g/mole

\text{Moles of }BaSO_4=\frac{0.403g}{233.38g/mole}=0.001727mole

Now we have to calculate the moles of barium ion.

From the balanced chemical reaction, we conclude that

As, 1 mole of barium sulfate produced from 1 mole of barium ion

So, 0.001727 mole of barium sulfate produced from 0.001727 mole of barium ion

Now we have to calculate the mass of barium ion.

\text{ Mass of }Ba^{2+}=\text{ Moles of }Ba^{2+}\times \text{ Molar mass of }Ba^{2+}

Molar mass of barium = 137.3 g/mol

\text{ Mass of }Ba^{2+}=(0.001727moles)\times (137.3g/mole)=0.2371g

Now we convert the mass of barium ion from gram to mg.

Conversion used : (1 g = 1000 mg)

Mass of barium ion = 0.2371 g = 237.1 mg

Now we have to calculate the mass percentage of barium in the compound.

Mass percent of barium = \frac{237.1mg}{441mg}\times 100=53.8\%

Thus, the mass percentage of barium in the compound is, 53.8 %

8 0
4 years ago
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