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babunello [35]
2 years ago
15

The reaction of a quantity of sulfur trioxide with water gives an acid with a volume of 2 l, containing 1.8 mol of the dissolved

oxide.
a)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.

b)Calculate the CM of the solution with a volume of 100 ml of ammonia (aq) to be added for complete neutralization of the acid solution.​
Chemistry
1 answer:
777dan777 [17]2 years ago
8 0

Answer :]

A.)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.

<em>Correct me if i'm wrong :]</em>

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dimulka [17.4K]

Answer:

I am so sorry I do not know this

Explanation:

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2 years ago
At 9°C a gas has a volume of 6.17 L. What is its volume when the gas is at standard temperature?
Alex17521 [72]

Answer:

V₂ = 5.97 L

Explanation:

Given data:

Initial temperature = 9°C (9+273 = 282 K)

Initial volume of gas  = 6.17 L

Final volume of gas = ?

Final temperature = standard = 273 K

Solution:

Formula:

The Charles Law will be apply to solve the given problem.

According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 6.17 L ×  273K /  282  k

V₂ = 1684.41 L.K / 282 K

V₂ = 5.97 L

5 0
3 years ago
Adding a catalyst to a system at equilibrium lowers the activation energy required by a system, which system, which shifts the e
GarryVolchara [31]

Answer: False

Explanation: Took the test

8 0
2 years ago
Read 2 more answers
The melting point of H₂O(s) is 0 °C. Would you expect the melting point of H₂S(s) to be 85 °C, 0 °C or -85 °C.? Justify your cho
dimulka [17.4K]

Answer:

-85 °C

Explanation:

O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.

H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter

molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.

5 0
2 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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