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4 years ago

For the process 2SO2(g) + O2(g) --> 2SO3(g),

Chemistry

1 answer:

CaHeK987 [17]4 years ago

8 0

Answer:

–187.9 J/K

Explanation:

The equation that relates the three quantities is:

$\Delta G = \Delta H - T \Delta S$

where

$\Delta G$ is the Gibbs free energy

$\Delta H$ is the change in enthalpy of the reaction

T is the absolute temperature

$\Delta S$ is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

$\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ$

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K

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When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many

Mumz [18]

Answer:

1.61 g Na₂S

Explanation:

To find the mass of sodium sulfide (Na₂S) generated from hydrogen sulfide (H₂S) and sodium hydroxide (NaOH), you need to (1) construct the balanced chemical equation, then (2) calculate the molar masses of each molecule involved, then (3) convert grams of each reagent to grams of the product (via the molar masses and mole-to-mole ratio from equation coefficients), and then (4) determine the limiting reagent and final answer. It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).

(Step 1)

The unbalanced equation:

H₂S + NaOH ---> Na₂S + H₂O

Reactants: 3 hydrogen, 1 sulfur, 1 sodium, 1 oxygen

Products: 2 hydrogen, 1 sulfur, 2 sodium, 1 oxygen

The balanced equation:

H₂S + 2 NaOH ---> Na₂S + 2 H₂O

Reactants: 4 hydrogen, 1 sulfur, 2 sodium 2 oxygen

Products: 4 hydrogen, 1 sulfur, 2 sodium, 4 oxygen

(Step 2)

Molar Mass (H₂S): 2(1.008 g/mol) + 32.065 g/mol

Molar Mass (H₂S): 34.081 g/mol

Molar Mass (NaOH): 22.990 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (NaOH): 39.998 g/mol

Molar Mass (Na₂S): 2(22.990 g/mol) + 32.065 g/mol

Molar Mass (Na₂S): 78.045 g/mol

(Step 3)

1.50 g H₂S 1 mole 1 mole Na₂S 78.045 g
----------------- x ---------------- x -------------------- x ----------------- =
34.081 g 1 mole H₂S 1 mole

= 3.43 g Na₂S

1.65 g NaOH 1 mole 1 mole Na₂S 78.045 g
-------------------- x ---------------- x ----------------------- x ---------------- =
39.998 g 2 moles NaOH 1 mole

= 1.61 g Na₂S

(Step 4)

Because NaOH generates less product, it will run out before all of the H₂S is used. This makes NaOH the limiting reagent and the final answer 1.61 grams Na₂S.

5 0

2 years ago

The wave like behavior of moving particles cannot be observed due to the.........of the particles wavelengths?

Nat2105 [25]

Answer is: 4)Mass

I really hope this helped you out!!!!

7 0

3 years ago

Read 2 more answers

Consider the following reaction. Zn(s) 2agno3(aq) --> 2ag(s) zn(no3)2(aq)when 16. 2 g of silver was produced, _____ mole(s) o

Crazy boy [7]

The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

W = given mass
M = molar mass

Moles of silver = 16.2g / 107.8g/mol = 0.15mol

From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.

Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g

Hence required mass of silver nitrate is 25.48g.

To know more about mass & moles, visit the below link:

brainly.com/question/19784089

#SPJ4

5 0

2 years ago

Strong acids are assumed 100% dissociated in water. True As a solution becomes more basic, the pOH of the solution increases. Fa

Kruka [31]

Answer:

Strong acids are assumed 100% dissociated in water- True

As a solution becomes more basic, the pOH of the solution increases- false

The conjugate base of a weak acid is a strong base- true

The Ka equilibrium constant always refers to the reaction of an acid with water to produce the conjugate base of the acid and the hydronium ion- True

As the Kb value for a base increases, base strength increases- true

The weaker the acid, the stronger the conjugate base- true

Explanation:

An acid is regarded as a strong acid if it attains 100% or complete dissociation in water.

The pOH decreases as a solution becomes more basic (as OH^- concentration increases).

Ka refers to the dissociation of an acid HA into H3O^+ and A^-.

The greater the base dissociation constant, the greater the base strength.

The weaker an acid is, the stronger , its conjugate base will be.

7 0

3 years ago

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

Theoretical yield = 3.51 g

3 0

3 years ago