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1. How many carbon atoms are in .76 mol of carbon atoms


2. 5.6 x 1022 particles, how many moles is that?


3. the mass in grams of 0.25 mol of water?


4.How many moles are in 48.2 g of calcium chloride (CaCl2)?


5. 11.2 L container of chlorine gas at standard temperature and pressure. How many moles of chlorine gas are in the container? 1 mol of gas in 22.4 L at standard temperature/ pressure.


6. the empirical formula of a compound that is 25.9% N and 74.1% O by mass.




answer any of each please also show how that would help

Respuesta :

anfabba15

Answer:

1. 4.57x10²³ atoms ; 2. 0.093 moles; 3. 4.5 grams ; 4. 0.434 moles , 5. 0.5 mol of Cl₂ ; 6. N₂O

Explanation:

1. 1 mol has NA atoms

0.76 mol will have (0.76 . 6.02x10²³) = 4.57x10²³ atoms

2. 6.02x10²³ particles are contained in 1 mol

5.6x10²² particles are contained in (5.6x10²²  . 1) / 6.02x10²³ = 0.093 moles

3. 1 mol of water weighs 18 g

0.25 mol . 18g/m = 4.5 grams

4. Molar mass of CaCl₂ = 110.98 g/m

Mol = Mass / molar mass → 48.2 g / 110.98 g/m = 0.434 moles

5. If the volume of Cl₂ at STP (as 1 mol of any gas) is 22.4L, and now we have the half of volume, we will have the half of moles, though. So, we have in the container 0.5 mol of Cl₂

6. 25.9% N and 74.1% O means that in 100 g of compound, we have 25.9 g of N and 74.1 g of O.

25.9 g of N / 14 g/m = 1.85 mol

74.1 g of O / 16 g/m = 4.63 mol

We divide the mol with the lowest value

4.63 mol / 1.85 mol = 2.5

1.85 mol / 1.85 mol = 1

N₂O

sebassandin

Answer:

1. [tex]4.73x10^{23}atoms C[/tex]

2. [tex]9.3x10^{-2}mol[/tex]

3. [tex]4.5gH_2O[/tex]

4. [tex]0.434molCaCl_2[/tex]

5. [tex]0.5molCl_2[/tex]

6.[tex]N_2O_5[/tex]

Explanation:

Hello,

In this case, considering the Avogadro's number relationship that states the following mathematical expression:

[tex]1mol=6.022x10^{23}particles[/tex]

In such a way, henceforth the solutions are shown:

1.

[tex]0.76molC*\frac{6.022x10^{23}atomsC}{1molC}=4.73x10^{23}atoms C[/tex]

2.

[tex]5.6x10^{22}particles*\frac{1mol}{6.022x10^{23}particles}=9.3x10^{-2}mol[/tex]

3.

[tex]0.25molH_2O*\frac{18gH_2O}{1molH_2O} =4.5gH_2O[/tex]

4.

[tex]48.2gCaCl_2*\frac{1molCaCl_2}{110.98gCaCl_2} =0.434molCaCl_2[/tex]

5. Considering the ideal gas equation:

[tex]n=\frac{PV}{RT}=\frac{1atm*11.2L}{0.082 \frac{atm*L}{mol*K}*273.15K}=0.5molCl_2[/tex]

6. In this case, one assumes those percentages as the masses of nitrogen and oxygen, in such a way, the resulting moles turn out:

[tex]n_N=\frac{25.9g}{14g/mol}=1.85\\ n_O=\frac{74.1g}{16g/mol} =4.63[/tex]

Next, diving by the nitrogen's moles, one obtain the subscripts as shown below:

[tex]N=\frac{1.85}{1.85}=1\\O=\frac{4.63}{1.85}=2.5[/tex]

Hence:

[tex]NO_{2.5}[/tex]

Finally, looking for the closest whole number, the empirical formula turns out:

[tex]N_2O_5[/tex]

Best regards.

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