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2 years ago

13

The last column of the periodic table contains the noble gases elements that do not easily form chemical bonds. Why don’t these

gases
tend to form chemical bonds?

1 answer:

ozzi2 years ago

6 0

Answer:

its is because the chemicals are very diffraction from each other so which makes it difficult to form a chemical bond.

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Cl2 + 2OH− → Cl− + ClO− + H2O

Nataly_w [17]

Cl2(g) -------> Cl-(aq) + ClO-(aq)

2e- + Cl2(g) -------> 2Cl-(aq) [reduction]

4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)

4 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

3 years ago

How many molecules are there in 21.4 grams of Ca(OH)2?

nata0808 [166]

The answer would be 0.288827452320528

4 0

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:

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

6 0

2 years ago

A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.5 kPa. If Po2 =

HACTEHA [7]

Answer:

A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.5 kPa.

<u>The pressure for oxygen is 3 kPa</u>

Explanation:

According to Dalton's Law of Partial Pressure total exerted by the mixture of non-reacting gases is equal to sum of the partial pressure of each gas.

$P_{total}=P_{1}+P_{2}+P_{3}$

So,

For , a gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure:

$P_{total}=P_{O_{2}}+P_{N_{2}}+P_{CO_{2}}$

$P_{total} = 32.5kPa$

$P_{O_{2}} = 6.5kPa$

$P_{N_{2}} = 23.0kPa$

Insert the values in :

$P_{total}=P_{O_{2}}+P_{N_{2}}+P_{CO_{2}}$

$32.5 kPa = 6.5 kPa + 23.0 kPa +P_{CO_{2}}$

$32.5 kPa = 29.5 kPa +P_{CO_{2}}$

$P_{CO_{2}}= 32.5 - 29.5$

$P_{CO_{2}}= 3kPa$

6 0

3 years ago