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3 years ago

11. We can do work on the air in the pump by compressing it. Initially, the gas has a volume of 200

Chemistry

1 answer:

natita [175]3 years ago

4 0

Answer:

The work done by the mixture during the compression is W=∫V2V1pdV. With the adiabatic condition of Equation 3.7. 14, we may write p as K/Vγ, where K=p1Vγ1=p2Vγ2.

Explanation:

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Two cubes sit side-by-side. The side length of a larger cube is 2.50 times the side length of the smaller cube. How many times l

nekit [7.7K]

Answer:

15.6 times longer is 15.625 times

Explanation:

the first one is 2.5 bigger, and then if you use that length on the area you can just cube 2.5

5 0

3 years ago

g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water

Marina CMI [18]

<u>Answer:</u> The boiling point of water in Tibet is 69.9°C

<u>Explanation:</u>

To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)

$P_2$ = final pressure = 240. mmHg

$\Delta H_{vap}$ = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature or normal boiling point of water = $100^oC=[100+273]K=373K$

$T_2$ = final temperature = ?

Putting values in above equation, we get:

$\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC$

Hence, the boiling point of water in Tibet is 69.9°C

3 0

3 years ago

Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges.

pshichka [43]

Three resonance structures can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

Learn more: brainly.com/question/4933048

7 0

3 years ago

A polymer P is made up of two monodisperse fractions; fraction A with molecular weight of 1000 g/mole and fraction B with a mole

vesna_86 [32]

Answer:

a)Number average molecular weight is 50, 500 g/mol

b) Weight average molecular weight is 99, 019.8 g/mol

Explanation:

We have a polymer P made up of two monodisperse fractions.

A with molecular weight of MA = 1000 g/mol and B with MB =100000 g/mol.

The batch contains an equal mole fraction of each component A and B.

Let's suppose a total number (Nt) of mols 2 moles. Equal fraction means XA = 0.5 and XB =0.5

Nt = 2 mol

Na = 2*0.5 = 1 mol

Nb = 2*0.5 = 1 mol.

So, we have 1 mol of A, 1 mol of B and 2 moles in total.

a) The number average molecular weight (NAM) is calculate using the mole numbers of each component. In this case, we will multiple each component molecular weight by the number of moles of each one. After that we will sum them and finally to divide by the total number of moles.

NAM = (Na*MA + Nb*MB)/(Nt)

NAM = (1 mol *1000 g/mol + 1*100000 g/mol ) /(2 mol)

NAM = 50500 g/mol

The number average molecular weight for the polymer P is 50,500 g/mol

b) Weight average molecular weight (WAM) is calculated using the mass quantities of each component. Weight mass of A (WA), weight mass of B (WB) are calculate using the moles of A, B and their molecular weights respectively. Total Weight (WT)

WA = Na*MA = 1 mol *1000 g/mol = 1000 g A

WB = Nb*MB = 1mol * 100000 g/mol = 100 000 gB

WT = WA + WB = 101 000 g

Now we will calculate average molecular using weights, we will multiple each component molecular weight by the mass of each one. After that we will sum them and finally to divide by the total mass.

WAM = (WA*MA + WB*MB)/(WT)

WAM = (1000 g *1000 g/mol + 100000 g*100000 g/mol )/(101 000 g)

WAM = 99 019.8 g/mol

The weight average molecular weight for polymer P is 99, 019.8 g/mol

6 0

4 years ago

During an experiment a thermometer was placed in a beaker containing hydrogen peroxide. The following observations were recorded

Rudiy27

Answer:

Chemical

Explanation:

The change was chemical, because it can no longer be returned to the original form. You cannot get back the bubbles or fizz from the air.

3 0

3 years ago

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