It is considered a DUI if the driver’s Blood Alcohol Content level is at or above .08.

Which of these can be determined based on the knowledge about weather?

PtichkaEL [24]

A: best time to go fishing at sea

6 0

3 years ago

Read 2 more answers

Please help! This is for a test and I want to have a good grade. Have a great day! :)

elena-s [515]

Answer:

genetic engineering !! :)

Explanation:

also, i love the pfp <3

6 0

3 years ago

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

3 years ago

Which physical property can be used to classify oxygen, helium, propane, and hydrogen as being similar?

PSYCHO15rus [73]

Answer:

state of matter

Explanation:

Because there are three state of matter namely:liquid,solid and gas and oxygen is a gas.

4 0

3 years ago

A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure

Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.

$P_i=P_t_o_t_a_l*X_i$

Where:

$P_i$ =Partial pressure

$P_t_o_t_a_l$ =Total pressure

$X_i$ =mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

moles of hydrogen gas

The molar mass of hydrogen gas ( $H_2$ ) is 2 g/mol, so: