2.20 moles Sn to grams

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

V₂ = 4,00L

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

V₂ = 16,00L

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

V₂ = 12,31L

I hope it helps!

6 0

4 years ago

Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.

skelet666 [1.2K]

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

4 0

3 years ago

Naming the hydrates?

slamgirl [31]

HEY mate here your answer

CoBr2 . 6 h2o = Cobalt(II) Bromide Hexahydrate

8 0

3 years ago

Read 2 more answers

If we add a catalyst to the following equation, CO + H2O + heat CO2 + H2, which way will the equilibrium shift?

Igoryamba

Answer:

No effect.

Explanation:

Hello,

In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.

Best regards.

7 0

3 years ago

How would you prepare 1.00 L of a 0.50 M solution of each of the following?

Lunna [17]

Answer:

Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.

Explanation:

Hello,

In this case, for the dilution process from concentrated 12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

$n_1=n_2$

Which can also be written in terms of concentrations and volumes:

$M_1V_1=M_2V_2$

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

$V_1=\frac{M_2V_2}{M_1} \\\\V_1=\frac{1.00L*0.5M}{12M}\\ \\V_1=0.0417L=41.7mL$

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.

Best regards.

6 0

3 years ago