The following reaction has an activation energy of 262 kJ/mol.

A solid precipitate is formed from the reaction of lead nitrate and sodium chromate.

Serhud [2]

I believe it's Filtration.

4 0

2 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

(c) $W_{isothermal}=5.743\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

2 years ago

1‑propanol ( n ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure a

Alex777 [14]

Answer:

$y_{prop}$ = 0.134; $y_{iso}$ = 0.866

The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr

Explanation:

For each of the solutions:

mole fraction of isopropanol ( $x_{iso}$ ) = 1 - mole fraction of propanol ( $x_{prop}$ ).

Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.

Furthermole, the partial pressure of isopropanol = $x_{iso}$ *vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr

The partial pressure of propanol = $x_{prop}$ *vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr

Similarly,

In the vapor phase,

The mole fraction of propanol ( $y_{prop}$ ) = $\frac{P_{prop} }{P_{prop}+P_{iso}}$

Where, $P_{prop}$ is the partial pressure of propanol and $P_{iso}$ is the partial pressure of isopropanol.

Therefore,

$y_{prop}$ = 5.26/(34.04+5.16) = 0.134

$y_{iso}$ = 1 - 0.134 = 0.866

5 0

3 years ago

Of the molecules below, only ________ is polar.a. SiCl4 b. CH4 c. CCl4 d. SeF4

ludmilkaskok [199]

Answer:

SeF4 is a polar molecule

Explanation:

SeF4 is a polar molecule because a polar molecule is any molecule that have lone pairs of electrons in the central atom or have atoms that are electronegative and the electrons between that are covalently bonded are not evenly distributed.

The electronegative atoms of flourine in SeF4 are not evenly distributed and kind pairs of electrons are on the central atom.

6 0

3 years ago

Determine the chemical formulas for the two compounds below. (Carbon, Hydrogen, and Oxygen).

tester [92]

Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.

1) CH3 - CH (OH) - CH (CH3) -CH3

2) CH3 - O - CH(CH3)-CH2 - CH3

I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -

4 0

3 years ago