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2 years ago

The process of passing electricity through water to chemically separate it into hydrogen and oxygen is called what?

Chemistry

2 answers:

SVEN [57.7K]2 years ago

3 0

electrolysis of water.

RSB [31]2 years ago

3 0

The answer is electrolysis
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.i hope that was helpful ;]

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How many molecules (or formula units) are in 138.56 g C4H10 Express your answer using four significant figures.

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2 years ago

Describe how to draw a 3-heptyne

lesantik [10]

You can't really describe it but this is what it looks like http://www.chemspider.com/Chemical-Structure.453291.html

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2 years ago

Two processes are described below: Process A: rapid moving water strikes a rock on the bottom of a stream and breaks it into pie

worty [1.4K]

Answer:

both represent physical weathering

Explanation:

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2 years ago

Read 2 more answers

The ionization of pure water forms _____.

Evgesh-ka [11]

Answer: The ionization of pure water forms hydroxide and hydronium ions.

Explanation:

Ionization is a reaction in the pure water in which water breaks down into its constituting ions that hydronium ion and hydroxide ions.

$H_2O+H_2O\rightleftharpoons H_3O^++OH^-$

One molecule of water looses its proton to form hydroxide ion and l=the lost protons get associated with another water molecule to form hydronium ion.

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2 years ago

The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen

tino4ka555 [31]

Answer: The mass percentage of $Tl_2SO_4$ is 5.86%

Explanation:

To calculate the mass percentage of $Tl_2SO_4$ in the sample it is necessary to know the mass of the solute ( $Tl_2SO_4$ in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the $TlI$ precipitate. We can establish a relation between the mass of $TlI$ and $Tl_2SO_4$ using the stoichiometry of the compounds:

$moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.$

Since for every mole of Tl in $TlI$ there are two moles of Tl in $Tl_2SO_4$ , we have:

$moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol$

Using the molar mass of $Tl_2SO_4$ we have:

$mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g$

Finally, we can use the mass percentage formula:

$mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%$

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2 years ago