All categories

2 years ago

Which statement is true about real gases with the correct reason:

Chemistry

2 answers:

victus00 [196]2 years ago

6 0

Volume of a real gas shows deviation from ideal gas on low temperature and high pressure

Only 2 options are correct then

Molecules have volume which causes deviations from ideal gases at low temperature.
Molecules have volume which causes deviations from ideal gases at high pressure.

LenKa [72]2 years ago

4 0

There are 2 possible answers here : b and d.

The Ideal Gas Equation is : PV = nRT

Here, when pressure is increased and temperature is lowered, the volume of the molecules will substantially decrease, which means it has deviated from ideal behavior.

You might be interested in

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Weight in grams of NaCl

Allisa [31]

Answer: 58.44g

Explanation: The molar mass of NaCl is 58.44g.

7 0

3 years ago

Read 2 more answers

What is formed when hydrobromic acid, hbr, and calcium hydroxide, ca(oh)2, are combined?

Lorico [155]

This is an acid – base reaction and this always result a salt and water in a neutralization reaction. 

The salt that is formed will be calcium bromide (calcium is located in group 2 so calcium bromide has a formula of CaBr2)

so essentially we got:

HBr + Ca(OH)2 ------> CaBr2 + H2O

balancing the elements: 

2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) + 2H2O(l)

6 0

3 years ago

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácid

Tanya [424]

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácido salicílico se convirtieron en 55.45 g de aspirina. ¿Cuál fue el porcentaje de rendimiento?

In English:

Aspirin is prepared by reacting salicylic acid with excess ethanoic anhydride. In one experiment, 50.05 g of salicylic acid was converted to 55.45 g of aspirin. What was the yield percentage?

Answer:

el rendimiento porcentual para la cantidad dada de ácido salicílico es 84.99 %

In English:

the percent yield for the given amount of salicylic acid is 84.99%

Explanation:

La ecuación química equilibrada para la reacción se puede escribir como:

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + HC₂H₃O₂

Para la reacción mostrada arriba; El reactivo limitante de la reacción es el ácido salicílico. Ahora; calcular el porcentaje de rendimiento; se espera que primero determinemos el rendimiento teórico de la reacción.

Entonces; la fórmula para calcular el porcentaje de rendimiento: $\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }$

El rendimiento teórico se determina de la siguiente manera:

50.05 g * 1 mol / 138.21 g / mol de C₇H₆O₃ * 1 mol de C₉H₈O₄ / 1 mol de C₇H₆O₃ * 180.157 g / mol de C₉H₈O₄ = 65.24 g de C₉H₈O₄

Porcentaje de rendimiento $\mathbf {= \frac{55.45 }{65.24 } *100 }$

Porcentaje de rendimiento = 84.99%

Por lo tanto, el porcentaje de rendimiento para la cantidad dada de ácido salicílico es 84.99%

In English:

The balanced chemical eqaution for the reaction can be written as:

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + HC₂H₃O₂

For the reaction shown above; The limiting reactant from the reaction is salicylic acid. Now; to calculate the percentage yield ; we are expected to first determine the theoretical yield of the reaction.

So; the formula for calculating the percentage yield $\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }$

The theoretical yield is determined as follows:

50.05 g * 1 mol/ 138.21 g/mol of C₇H₆O₃ * 1 mol of C₉H₈O₄/ 1 mol of C₇H₆O₃ * 180.157 g/mol of C₉H₈O₄ = 65.24 g of C₉H₈O₄ is produced

Percentage yield $\mathbf {= \frac{55.45 }{65.24 } *100 }$

Percentage yield = 84.99%

Thus, the percent yield for the given amount of salicylic acid is 84.99%

7 0

3 years ago

The highest parts of a transverse wave are called a. troughs. b. crests. c. nodes. d. wavelengths.

Zepler [3.9K]

The highest is b crest.

5 0

3 years ago

Read 2 more answers