What three characteristics of matter does chemistry deal with

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago

8. Neil pogo sticks from his locker to his science class. He travels 8 m east then 8 m west

jeyben [28]

Answer:

12

Explanation:

5 0

3 years ago

Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3N O 2 − + 8H + + C r 2 O 7 2− → 3N O 3 −

Fantom [35]

Answer:

NO2- is the reducing agent.

Cr2O7_2- is the oxidizing agent.

H+ is neither

Explanation:

Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.

Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.

Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.

Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.

The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.

4 0

4 years ago

What is the value of Kw at 25°C? 1.0 x 10–14 1.0 x 10–7 1 7 14

romanna [79]

1.0 x 10-14. That is the value of Kw at 25 degrees C.

3 0

3 years ago

Read 2 more answers

The density of nitric oxide (NO) gas at 0.866 atm and 46.2 degrees celcius is ____ g/L

Artyom0805 [142]

Answer:

d = 0.992 g/L

Explanation:

Data Given:

Pressure of nitric oxide (NO) = 0.866 atm

Temperature of a gas = 46.2° C

Convert the temperature to kelvin = 46.2° C + 273

temperature in kelvin = 319.2 K

density of nitric oxide (NO) = ?

Solution:

Density of a gas can be calculated by

d = PM /RT

Where

d = density

P = Pressure

M = molar mass of gas

R = ideal gas constant = 0.0821 L atm mol⁻¹ K⁻¹

T = temperature

So,

Molar mass of NO = 30 g/mol

Put values in the formula:

d = PM /RT

d = 0.866 atm × 30 g/mol / 0.0821 L atm mol⁻¹ K⁻¹ × 319.2 K

d = 25.98 atm. g/mol / 26.2 L atm mol⁻¹

d = 0.992 g/L

8 0

3 years ago