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3 years ago

Consider the reaction:

1 answer:

8 0

Gosh, I love chem!

[H2O]/t should be 0.085M/s

this is because of the coefficients in front of H2S and H2O being the same.

the significance of this answer is that the H2S disappears at the same rate H2O appears.

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Which Of These Periods Contain Elements With Electrons s,p,d and f orbitals? Periods 1-3 Periods 1-4

Elanso [62]

The answer would be periods 6-7 :)

7 0

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:

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

6 0

2 years ago

Brainliest for correct answer please show all work

Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

Question 1:

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

Balance the number of Na:

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

Balance the number of K:

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

The equation is now balanced.

Question 2:

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

Balance number of Li and F atoms:

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

The equation is now balanced.

Question 3:

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

Balance the number of P atoms:

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

Balance the number of O atoms:

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

The equation is now balanced.

4 0

2 years ago

2 Calculate What is the density of a liquid with a mass of 17.4g and a volume of 20 mL?

jolli1 [7]

Answer:

P= 0.87g/mL or 0.87g/cm^3

Explanation:

P=m/v

P=density

P=17.4g/20mL

P= 0.87g/mL

1mL=1cm^3

8 0

3 years ago

What element forms an ion with an electronic configuration of [kr] and a –2 charge?

nekit [7.7K]

Must be Selenium (Se). It is two space away from Kr, which means that it needs two extra electron to be like a noble gas, Kr.

6 0

2 years ago

Read 2 more answers