Wjat is 0.06m to cm

A chemical reaction has reached equilibrium when ________. all products have been removed from the reaction mixture the catalyst

soldier1979 [14.2K]

rate of forward reaction

5 0

2 years ago

Read 2 more answers

I need please fast ASAP

sdas [7]

Kindly checkout the attached pic. It contains all your required questions.

4 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

134 grams of nitric acid is added to 512 grams of water. Calculate the molality of nitric acid.

Nana76 [90]

Answer: 4.15234 m

512 g H2O * $\frac{1 kg}{1000 g}$ = 0.512 kg H2O

Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol

H = 1.008 g/mol

N = 14.007 g/mol

O3 = 3*15.999

134 g HNO₃ * $\frac{mol}{63.012 g}$ = 2.126 mol

m = $\frac{2.126 mol}{0.512 kg}$ = 4.15234 m

6 0

3 years ago

Read 2 more answers

If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present?if so, what mass of liquid?

Anastasy [175]

Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. 

Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol 

Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. 

Then : 0.05x 137.5 = 6.88gm of vapor 

If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d

8 0

2 years ago

Wjat is 0.06m to cm​

Wjat is 0.06m to cm