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Anit [1.1K]
2 years ago
11

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

, what is the pressure in the flask after reaction is complete?
Chemistry
1 answer:
diamong [38]2 years ago
5 0

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 25^oC=273+25=298K

V = volume = 4.0 L

n_1 = moles of SO_2 = 0.20 mol

n_2 = moles of O_2 = 0.20 mol

Now put all the given values in the above expression, we get:

P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}

P=2.4atm

Thus, the pressure in the flask after reaction complete is, 2.4 atm

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solong [7]

Answer:

The mass is recorded as 32.075 g

Explanation:

"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the  whole figure.

8 0
2 years ago
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A gas occupies a volume of 72 ml at 400 k and 800 torr. if the temperature drops to 200 k and the pressure changes to 400 torr,
satela [25.4K]
We are tasked to solve for the volume of the gas that occupies when pressure and temperature changes to 400 Torr and 200 Kelvin from Torr and 400 Kelvin. We can use ideal gas law assuming constant gas composition and close system. The solution is shown below:
P1V1 / T1 = P2V2 / T2
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The answer for the volume is 72 ml.
7 0
2 years ago
the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many
8_murik_8 [283]

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or  291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

 mass remains = 4g

Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

8 0
2 years ago
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In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as
lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio  of any nuclei ranges from  to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the  Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons =  Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons  =  Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons  =  Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons  =  Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons  =  Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons  =  Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons  =  Neutron- proton ratio = N/P =  7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons =  Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0
2 years ago
A solid metal oxide crystallizes in a cubic unit cell. In the unit cell there are metal (M) ions on every corner and in the cent
kotegsom [21]

Answer:

The empirical formula of the solid metal oxide is : MO

Explanation:

M atom is present in the every corner and in the center of every face.

Number of m atoms :

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Number of tetrahedral voids in F.C.C = 2n = 2 × 4 = 8

Oxide ion is present in the half of the tetrahedral void

Number oxide ions = \frac{8}{2}=4

The molecular formula of the solid metal oxide is : M_4O_4=MO

The empirical formula represent the lowest number of atoms present in a compound.

The empirical formula of the solid metal oxide is : MO

8 0
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