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sergij07 [2.7K]
2 years ago
6

What are examples of solid solutions?

Chemistry
1 answer:
12345 [234]2 years ago
3 0
Some examples are the wall, wood, tv's, desk, metal, doors, and books
You might be interested in
what pressure is generated when 5 mol of ethane is stored in a volume of 8 dm 3 at 25°C? Base calculations on each of the follow
Goshia [24]

Answer:

a) 40,75 atm

b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

                                     PV = nRT

where:

  • P = pressure [atm]
  • V = volume [L]
  • n = number of mole of gas [n]
  • R= gas constant = 0,08205 [atm.L/mol.°K]
  • T=absolute temperature [°K]

<em>Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.</em>

The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the <u>intermolecular forces</u> and the <u>space occupied</u> by the gas

\frac{Pv}{RT} = 1 + \frac{B}{v}

where:

  • v is the molar volume [L/mol]
  • B is the second virial coefficient [L/mol]
  • P the pressure [atm]
  • R the gas constant = 0,08205 [atm.L/mol.°K]

a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}

and v = 3 L/5 moles = 0,6 L/mol

P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm

5 0
3 years ago
Calculate the mean of 2, 5 and 3. write your answer to 2 decimal places. please help me on this
makvit [3.9K]

Hello.

The answer is: 3.3333

To get the answer add 2,5 and 3 that is 10 then divide by 3 to get 3.3

Have a nice day

6 0
3 years ago
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0
3 years ago
A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s
s2008m [1.1K]

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=1.0atm\\V_1=3L\\P_2=3.5atm\\V_2=?

Putting values in above equation, we get:

1.0\times 3L=3.5\times V_2\\\\V_2=0.86

Thus the squirrel's balloon will be 0.86 L

7 0
2 years ago
Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water
barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same  

The dilution formula

 M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M

5 0
3 years ago
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