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3 years ago

PLEASE HELP ASAP!!!!!!!!

Chemistry

1 answer:

Step2247 [10]3 years ago

3 0

Answer:

1).There are 6.87 x 1023 atoms in 1.14 mol SO3, or sulfur trioxide (mol is the abbreviation for mole)

2).6.022*1023 R.P . is equal to 144

3)7.02 g Si

4)104*10 raised to 1

5)idk

6)142.04 g/mol

7)atom

8) idk

Explanation:

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Lines on a map joining places that have the same temperature?

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Answer:

Isotherms

Explanation:

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3 years ago

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Determine the bonding type for boron trihydride given the electronegativity on the Pauling scale for boron is 2.0 and for hydrog

Anika [276]

Answer: The bond between boron and hydrogen in boron trihydride is covalent bond.

Explanation:

The type of bonding between the atoms forming a compound is determined by using the electronegativity difference between the atoms. According to the pauling's electronegativity rule:

If $\Delta \chi=0$ , then the bond is non-polar.
If $\Delta \chi\leq 1.7$ , then the bond will be covalent.
If $\Delta \chi>1.7$ , then the bond will be ionic.

We are given:

Electronegativity for boron = 2.0

Electronegativity for hydrogen = 2.1

$\Delta \chi=\chi_{H}-\chi_{B}\\\\\Delta \chi=2.1-2.0=0.1$

As, $\Delta \chi$ is less than 1.7 and not equal to 0. Hence, the bond between boron and hydrogen is covalent bond.

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3 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:

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