Ask question

Ask question

All categories

sergij07 [2.7K]

2 years ago

15

A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres

sure is 1.350 atm . Part A What is the partial pressure of argon, PAr, in the flask?

1 answer:

tatiyna2 years ago

4 0

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 $m^{3}$

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 $\frac{J}{kg k}$

From ideal gas equation P V = m RT

⇒ P = $\frac{m R T}{V}$

Put all the values in above formula we get

⇒ P = $\frac{0.00115}{0.001}$ × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa

You might be interested in

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l)

Marina CMI [18]

Answer:

Please see the complete formt of the question below

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)

According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.

The answer to the above question is

The limiting reactant is the MnO₂

Explanation:

To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂

Molar mass of MnO₂ = 86.9368 g/mol

Molar mass of HCl = 36.46 g/mol

From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂

However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed

7 0

2 years ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

2 years ago

Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right

svetlana [45]

Answer:

B = CHCl2 + Cl2 --> CHCl3 + Cl

Explanation:

Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.

Free radical chlorination is divided into 3 steps which are:

The initiation step

The propagation step

The termination step

So in reference to the question, propagation step involves two steps.

The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.

The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.

You would find in the attachment the 2 step mechanism.

3 0

2 years ago

How many hydrogen molecules are there in 1 ton of hydrogen?

zaharov [31]

Hydrogen gas(H2) has a molar mass of 2 g. Molar mass of a substance is defined as the mass of 1 mole of that substance. And by 1 mole it is meant a collection of 6.022*10^23 particles of that substance.

So number of moles of H2 are 0.5 in this case. And thus it means there are (6.022*10^23)*0.5 particles( here they are molecules) in 1g of H2.

6 0

2 years ago

A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. Th

Setler [38]

Answer: (i)The object is at a distance of 120cm from the lens while the image is at a distance of 160cm from the lens. (ii)The image is Real

Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Explanation:

The required formulas are;

1. lens formula given by, <em>1/s' + 1/s = 1/f</em>

2. magnification = <em>h'/h = s'/s</em>

where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.

h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)

solving for s' and s in eqn (2)

4.50/3.20=s'/s

1.4=s'/s

s'=1.4s

to find the values of s' and s, we use equation (1) and substitute s' = 1.4s

1/1.4s + 1/s = 1/70

2.4/1.4s =1/70

s = 2.4*70/1.4 = 120cm

s' = 1.4*120 = 160cm

Therefore, the object is on the left, 120cm from the lens while the image is 160cm on the right of the lens

<em>Since the object is at a distance between f and 2f from the lens, the image formed is</em> real, inverted and magnified

3 0

2 years ago